NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation

A quadratic equation In the variable x Is the equation of the form ax²+bx+c=0, where a, b, c are real numbers, a ≠ 0. For example.,

1. 3x² + 5x -1=0
2. 3x-x² + 1 =0

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Roots Of Quadratic Equation

A real number a is called a root of the quadratic equation ax² + bx + c = 0, a= 0 if
aα² + bα + c = 0

i.e., x = α satisfies the equation ax² + bx + c = 0

or x = α is a solution of the equation ax² + bx + c = 0

The roots of a quadratic equation ax² + bx + c = 0 are called zeroes of the polynomial ax² + bx + c.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Solution Of A Quadratic Equation By Factorisation Method

Consider the quadratic equation ax² + bx + c = 0, a≠0

Let it be expressed as a product of two linear expressions (factors) namely (px + q) and (rx + s) where p, q, r, s are real numbers and p ≠ 0, r ≠ 0, then

ax² + bx + c = 0 ⇒ (px + q)(rx + s) = 0

⇒ px + q = 0  or  rx + s = 0

Read and Learn More Class 10 Maths Solutions Exemplar

⇒ \(x=-\frac{q}{p}\)  or  \(x=-\frac{s}{r}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Solution Of A Quadratic Equation Solved Examples

Question 1. Which of the following are the solutions of 2x² – 5x – 3 = 0?

1. x = 2
2. x = 3
3. \(x=-\frac{1}{2}\)

Solution:

The given equation is 2x² – 5x – 3 = 0

1. On substituting x = 2 in the given equation

L.H.S. = 2×2² – 5×2-3 = 8-10-3 = -5 ≠ R.H.S.

∴ x = 2 is not a solution of 2x² – 5x – 3 = 0

2. On substituting x = 3 in the given equation

L.H.S. = 2×3²– 5×3-3 = 18 -15- 3 = 0 = R.H.S. 2

∴ x = 3 is a solution of 2ײ – 5x – 3 = 0

3. On substituting x = \(-\frac{1}{2}\) in the given equation

L.H.S = \(2 \times\left(\frac{-1}{2}\right)^2-5 \times\left(\frac{-1}{2}\right)-3\)

⇒ \(2 \times \frac{1}{4}+5 \times \frac{1}{2}-3\)

⇒ \(\frac{1}{2}+\frac{5}{2}-3=\frac{1+5-6}{2}\)

=0 = R.H.S

∴ x = \(-\frac{1}{2}\) is a solution of 2×2-5x-3 = 0

Question 2. If x = 2 and x = 3 are roots of the equation 3x²-mx+ 2n = 0, then find the values of m and n.

Solution:

Since, x = 2 is a solution of 3x² – mx + 2n = 0

∴ 3 × (2)² -m×2 + 2n = 0

⇒ 12 -2m + 2n=0

⇒ -m + n = -6 ……(1)

Again, x = 3 is a solution of 3x² – mx + 2n = 0

∴ 3 × (3)² – m × 3 + 2n = 0

⇒ 27 – 3m + 2n = 0

⇒ -3m + 2n = -27 ……(2)

On multiplying equation (1) by 2 and subtracting from (2) we get

-m =-15 or m= 15

On substituting the value of m in equation (1) we get

-15 + n =-6

⇒ n= -6+15

⇒ n=9

Hence, The values of m = 15 and n = 9

Question 3. Solve the following quadratic equation : (3x-5)(2x + 3) = 0

Solution:

Given equation is (3x – 5)(2x + 3) = 0

⇒ 3x -5=0 or 2x + 3 = 0

⇒ 3x = 5 or 2x = -3

⇒ \(x=\frac{5}{3}\)  or  \(x=-\frac{3}{2}\)

Here, x = \(\frac{5}{3}\) and x = \(-\frac{3}{2}\) are the solutions.

Question 4. Find the roots of the following quadratic equation by factorization: 2z² + az – a² = 0

Solution:

Given equation is 2z² + az – a² =0

⇒ 2z² + (2a – a) z -a² = 0

⇒ 2z² + 2az – az-a² = 0

⇒ 2z(z + a) -a(z + a) = 0

⇒ (z + a)(2z – a) = 0

⇒ z + a = 0 or 2z-a = 0

when z + a = 0 ⇒ z = – a

and 2z – a = 0 ⇒ z = \(\frac{a}{2}\)

Hence, the roots of the equation are -a and \(\frac{a}{2}\)

Question 5. Solve the following quadratic equations by factorization:

1. 4-11x = 3x²
2. \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Solution:

1. Given equation is 4 -11x = 3x²

⇒ 3x²+11x-4 = 0

⇒ 3x²+(12-1)-4 = 0

⇒ 3x²+12x-x-4 = 0

⇒ 3x(x+4)(x+4)

⇒ (3x-1)(x+4)=0  or x+4 = 0

when 3x+1 = 0

⇒ \(x=\frac{1}{3}\)

and x + 4 = 0

x = -4

Hence,\(\frac{1}{3}\) and- 4 are roots of equation

2. Given equation is \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Multiplying both sides by 8, we get

⇒ 8x²– 22x + 15 = 0

⇒ 8x² – (12 + 10)x+ 15 = 0

⇒ 8x²– 12x- 10x+ 15 = 0

⇒ 4x(2x- 3) – 5(2x- 3) = 0

⇒ (2x-3)(4x-5) = 0

∴ either 2x- 3 = 0  or 4x- 5 = 0

2x = 3 or 4x = 5

⇒ \(x=\frac{3}{2}\) or \(x=\frac{5}{4}\)

Hence \(x=\frac{3}{2}\) and \(x=\frac{5}{4}\) are the roots of given equation.

Question 6. Solve the following quadratic equation :

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

Solution:

Given equation is

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

⇒ \(x^2-x-\sqrt{2} x+\sqrt{2}=0\)

⇒ \(x(x-1)-\sqrt{2}(x-1)=0\)

⇒ \((x-1)(x-\sqrt{2})=0\)

x-1 = 0 or \(x-\sqrt{2}=0\)

when x-1 = 0 ⇒ x = 1

and \(x-\sqrt{2}=0\) ⇒ \(\sqrt{2}\)

Hence, 1 and \(\sqrt{2}\) are roots of the equation.

Question 7. Solve the following quadratic equation: a²b²x² + b²x- a²x-1 = 0

Solution:

Given equation is

a²b²x² + b²x- a²x-1=0

b²x(a2x + 1)-1 (a²x + 1 ) = 0

(a²x+1)(b²x- 1) = 0

a²x + 1 = 0 or b²x-1=0

when a²x+1=0 ⇒ \(x=-\frac{1}{a^2}\)

and b²x- 1=0 ⇒ \(x=\frac{1}{b^2}\)

Hence, \(-\frac{1}{a^2} \text { and } \frac{1}{b^2}\) are roots equation.

Question 8. Solve the following quadratic equation: 4x²– 2(a² + b²)x + a²b²=0

Solution.

Given equation is

4x² -2(a² + b²)x + a²b² = 0

4x²– 2a²x- 2b²x + a²b² = 0

2x(2x- a²)-b²(2x -a²) = 0

(2x-2)(2x-b²) = 0

2x-a² = 0

or 2x-b² = 0

when 2x-a² = 0 ⇒ \(x=\frac{a^2}{2}\)

and 2x-b² = 0 ⇒ \(x=\frac{b^2}{2}\)

Question 9. Solve the following equation :

⇒ \(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3},(x \neq 4,3)\)

Solution:

⇒ \( \frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\)

⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)

⇒ \(\frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}\)

⇒ \(\frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}\)

⇒ \(3\left(4 x^2-19 x+20\right)=25\left(x^2-7 x+12\right)\)

⇒ \(12 x^2-57 x+60=25 x^2-175 x+300\)

⇒ \(25 x^2-175 x+300-12 x^2+57 x-60=0\)

⇒ \(13 x^2-118 x+240=0\)

⇒ \(13 x^2-(78+40) x+240=0\)

⇒ \(13 x^2-78 x-40 x+240=0\)

⇒ 13x(x-6)-40(x-6) = 0

⇒ (x-6)-409x-6) = 0

⇒ x-6 = 0 or 13x-40 = 0

⇒ x-6 0  or \(x=\frac{40}{13}\)

Hence, 6 and \(\frac{40}{13}\) are roots of the equation.

Question 10. Solve the following equation:

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5,(x \neq-3,1)\)

Solution:

Given equation is

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5\)…..(1)

Let \(\frac{2 x-1}{x+3}=y\)

Hence, \(\frac{x+3}{2 x-1}=\frac{1}{y}\)

Now from equation (1)

⇒ \(2 y-\frac{3}{y}=5\)

⇒ \(2 y^2-3=5 y\)

⇒ \(2 y^2-5 y-3=0\)

⇒ \(2 y^2-(6-1) y-3=0\)

⇒ \(2 y^2-6 y+1 y-3=0\)

⇒ 2y(y-3)+1(y-3)=0

⇒ (2y+1)(y-3)=0

⇒ 2y+1=0 or y-3=0

when 2y + 1 = 0 ⇒ y = \(-\frac{1}{2}\)

and y-3 = 0 ⇒ y = 3

Substituting values of y in equation (2)

when y = \(-\frac{1}{2}\)

⇒ \(\frac{2 x-1}{x+3}=-\frac{1}{2}\)

⇒ 2(2x-1) = -1(x + 3)

⇒ 4x- 2 = -x- 3

⇒ 5x = -1

⇒ x = \(-\frac{1}{5}\)

when y = 3

⇒ \(\frac{2 x-1}{x+3}=3\)

⇒ 2x- 1 = 3(x + 3) ⇒ 2x- 1 = 3x+9

⇒ -x = 10        ⇒           x =-10

Hence, x = -10 or x = \(-\frac{1}{5}\) are roots of the equation.

Question 11. Solve the equation:

⇒\(\frac{a}{x-b}+\frac{b}{x-a}=2 \quad(x \neq b, a)\)

Solution:

Given equation is \(\frac{a}{x-b}+\frac{b}{x-a}=2\)

⇒ \(\frac{a}{x-b}+\frac{b}{x-a}=1+1\)

⇒ \(\frac{a}{x-b}-1+\frac{b}{x-a}-1=0\)

⇒ \(\frac{a-x+b}{x-b}+\frac{b-x+a}{x-a}=0\)

⇒ \((a+b-x)\left(\frac{1}{x-b}+\frac{1}{x-a}\right)=0\)

a+b -x= 0 or \(\frac{1}{x-b}+\frac{1}{x-a}=0\)

when a +b-x= 0 ⇒  x=a +b

and when  \(\frac{1}{x-b}+\frac{1}{x-a}=0\) ⇒ \(\frac{x-a+x-b}{(x-b)(x-a)}=0\)

⇒ 2x-a-b = 0

⇒ 2x= a + b

⇒ \(x=\frac{a+b}{2}\)

Alternatively,

⇒ \(\frac{1}{x-b}=-\frac{1}{x-a}\)

⇒ \(x-b=a-x\)

⇒ \(2 x=a+b\)

⇒ \(x=\frac{a+b}{2}\)

Hence, x = a + b and \(x=\frac{a+b}{2}\) arc roots of the equation.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Formula

The roots of the quadratic equation ax² + bx+ c= 0 where a ≠ 0 can be obtained by using the formula.

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Proof: Given ax2 + bx + c = 0

Dividing each term by a, we get

⇒ \(x^2+\frac{b}{a} x+\frac{c}{a}=0\)   (∵ a ≠ 0)

⇒ \(x^2+\frac{b}{a} x=-\frac{c}{a}\)  (transposing the constant)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e. }\left(\frac{b}{2 a}\right)^2\) on both sides, we get

⇒ \(x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\left(\frac{b}{2 a}\right)^2-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)

Taking square root on both sides, we get

⇒ \(x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)  (try to understand ‘±’)

⇒ \(x=\frac{-b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\) or ⇒ \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

This is known as “Sridharacharya Formula” or “quadratic formula”.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Discriminant

For the quadratic equation ax² + bx + c = 0, the expression D = b²– 4ac is called the discriminant. Roots of ax² + bx + c = 0 are real, only when b² – 4ac > 0, otherwise they are imaginary (not real).

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Sum Of Roots And Product Of Roots

We know that the two roots of a quadratic equation ax² + bx + c = 0 are

⇒ \(\alpha=\frac{-b+\sqrt{D}}{2 a} \quad \text { and } \quad \beta=\frac{-b-\sqrt{D}}{2 a}\) where D=b2- 4acis called the discriminant.

∴ Sum of roots :

⇒ \(\alpha+\beta=\frac{-b+\sqrt{D}}{2 a}+\frac{-b-\sqrt{D}}{2 a}=\frac{-b+\sqrt{D}-b-\sqrt{D}}{2 a}=\frac{-2 b}{2 a}=\frac{-b}{a}\)

∴ Sum of roots = \(\frac{-b}{a}=-\frac{\text { Coeff. of } x}{\text { Coeff. of } x^2}\)

Product of roots :

⇒ \(\alpha \beta=\frac{-b+\sqrt{D}}{2 a} \times \frac{-b-\sqrt{D}}{2 a}\)

⇒ \(\frac{(-b)^2-(\sqrt{D})^2}{4 a^2}=\frac{b^2-D}{4 a^2}=\frac{b^2-\left(b^2-4 a c\right)}{4 a^2}=\frac{4 a c}{4 a^2}=\frac{c}{a}\)

∴ Product of roots : \(\frac{c}{a}=\frac{\text { Constant term }}{\text { Coeff. of } x^2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Sum Of Roots And Product Of Roots Solved Examples

Question 1. Find the roots of the following quadratic equation, if they exist by the method of completing the square.

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Solution:

Given equation is

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Dividing both sides by 3

⇒ \(x^2+\frac{4 \sqrt{3}}{3} x+\frac{4}{3}=0\)

⇒ \(x^2+\frac{4}{\sqrt{3}} x=\frac{-4}{3}\)

Adding \(\left(\frac{\text { coefficient of } x}{2}\right)^2 \text { i.e., }\left(\frac{2}{\sqrt{3}}\right)^2=\frac{4}{3}\) on both sides

⇒ \(x^2+\frac{4}{\sqrt{3}} x+\frac{4}{3}=-\frac{4}{3}+\frac{4}{3}\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)^2=0\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)=0 \quad \text { and } \quad\left(x+\frac{2}{\sqrt{3}}\right)=0\)

⇒ \(x=\frac{-2}{\sqrt{3}} \quad \text { and } \quad x=\frac{-2}{\sqrt{3}}\)

Hence, roots of the equation are \(\frac{-2}{\sqrt{3}} \text { and } \frac{-2}{\sqrt{3}}\)

Question 2. Find roots of the following quadratic equations by using the quadratic formula, if they exist.

1. 3x² +x-4 = 0
2. 3x² +x+ 4 = 0

Solution:

1. Given equation is 3x² +x-4 = 0

On comparing with ax² + bx+ c = 0, we get

a = 3, b = 1 and c = -4

∴ Discriminant, D = b² – 4ac

D = (1)² – 4 × 3 × (-4)

D = 1 +48

D = 49 > 0

Hence, the given equation has two real roots.

∴ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{49}}{6}=\frac{-1 \pm 7}{6}=\frac{6}{6}\)

⇒ \(\text { or } \frac{-8}{6}\)

⇒ \(1 \text { or } \frac{-4}{3}\)

⇒ \(x=1, \frac{-4}{3}\) are roots of the equation.

2. Given equation is 3x² +x+ 4 = 0

On comparing with ax² + bx + c = 0, we get

a = 3, b = 1 and c = 4

∴ Discriminant, D – b² – 4ac

D = (1)² – 4 × 3 × 4

D = 1 – 48

D = -47 < 0

Hence, the equation has no real roots.

Question 3. Find roots of the equation by quadratic formula : x+x -(a + 2) (a+1) = 0
Solution:

The given equation is x² + x- (a + 2) (a +1) = 0

Comparing it with Ax² + Bx + C = 0, we get

A = 1, B = 1 and C = -(a + 2) (a+1)

∴ \(x=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)

⇒ \(x=\frac{-1 \pm \sqrt{1^2-4 \times 1 \times[-(a+2)(a+1)]}}{\cdots}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4\left(a^2+3 a+2\right)}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4 a^2+12 a+8}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{4 a^2+12 a+9}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{(2 a+3)^2}}{2}\)

⇒ \(x=\frac{-1 \pm(2 a+3)}{2}\)

⇒ \(x=\frac{-1+2 a+3}{2} \text { and } \frac{-1-2 a-3}{2}\)

⇒ \(x=\frac{2 a+2}{2} \text { and } \frac{-2 a-4}{2}\)

⇒ x = (a + 1) and -(a + 2) are roots of the equation.

Question 4. Solve the following equation by the method of completing the square :

⇒ \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Solution:

We have, \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Dividing each term by \(4 \sqrt{3}\) we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x-\frac{2 \sqrt{3}}{4 \sqrt{3}}=0\)

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x=\frac{1}{2}\)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e., }\left(\frac{5}{8 \sqrt{3}}\right)^2\) to both sides, we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x+\left(\frac{5}{8 \sqrt{3}}\right)^2=\left(\frac{5}{8 \sqrt{3}}\right)^2+\frac{1}{2}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25}{192}+\frac{1}{2} \Rightarrow\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25+96}{192}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{121}{192}\)

Taking the square root of both sides, we get

⇒ \(x+\frac{5}{8 \sqrt{3}}= \pm \frac{11}{8 \sqrt{3}}\)

∴ \(x=-\frac{5}{8 \sqrt{3}} \pm \frac{11}{8 \sqrt{3}}=\frac{-5 \pm 11}{8 \sqrt{3}}\)

⇒ \(x=\frac{-5+11}{8 \sqrt{3}} \quad \text { or } \quad x=\frac{-5-11}{8 \sqrt{3}}\)

⇒ \(x=\frac{3}{4 \sqrt{3}} \quad \text { or } \quad x=\frac{-2}{\sqrt{3}}\)

Hence x = \(\frac{3}{4 \sqrt{3}} \text { or } x=\frac{-2}{\sqrt{3}}\) are the solutions of given equation.

Question 5. Solve: x² + x- (a + 2) (a + 1) = 0 by

1. Factorisation
2. Method of completing the square

Solution:

1. By factorisation :

We have x² +x- (a + 2) (a + 1) = 0

⇒ x²+x× 1 -{a + 2) (a+ 1) = 0

⇒ x² +x[(a + 2) — (a + 1)] — (a + 2) (n + 1) = 0

⇒ x² +x(a + 2) -x(a + 1) – (a + 2) (a + 1) = 0

⇒ x[x+ (a + 2)] – (a + 1)[x+ (a + 2)] = 0

⇒ [x+(a + 2)] [x-(a+ 1)] = 0

∴ either x+ (a + 2) = 0  or x- (a + 1) = 0

⇒ x = -(a + 2) or x=(a+ 1)

2. By the method of completing the square:

We have x² + x- (a + 2) (a + 1) = 0

⇒ x² +x=(a + 2) (a + 1)

Adding \(\left(\frac{1}{2}\right)^2\) on both sides, we get

⇒ \(x^2+x+\left(\frac{1}{2}\right)^2=a^2+3 a+2+\left(\frac{1}{2}\right)^2\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\frac{4 a^2+12 a+9}{4}\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2 a+3}{2}\right)^2\)

Taking the square root of both sides, we get

⇒ \(x+\frac{1}{2}= \pm \frac{2 a+3}{2}\)

∴ \(x=\frac{-1}{2} \pm \frac{2 a+3}{2}=\frac{-1 \pm(2 a+3)}{2}\)

∴ \(x=\frac{-1+2 a+3}{2}\text { or }x=\frac{-1-(2 a+3)}{2}\)

⇒ \(x=\frac{2(a+1)}{2}\text { or }x=\frac{-2(a+2)}{2}\)

⇒ \(x=(a+1)\text { or }x=-(a+2)\)

Question 6. Let f(x) = 3x² – 5x- 1. Then solve f(x) = 0 by

1. Factoring the quadratic
2. Using the quadratic formula
3. Completing the square and then rewrite f(x) in the formv4(x±B)² ± C.

Solution:

f(x) = 3x²– 5x- 1

f(x) = 0

3x²-5x- 1=0

1. The given quadratic equation cannot be fully factorised using real integers. So, it is better to solve this equation by any other method.

2. 3x-5x- 1 =0

Compare it with ax² +bx + c = 0, we get

a = 3, b = -5,c =-1

∴ Let two roots of this equation are

⇒ \(\alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)+\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\frac{5+\sqrt{37}}{6}\)

and

⇒ \(\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)-\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\beta=\frac{5-\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} { and } \frac{5-\sqrt{37}}{6}\)

3. \(3 x^2-5 x-1=0\)

⇒ \(x^2-\frac{5}{3} x-\frac{1}{3}=0\)

⇒ \(x^2-\frac{5}{3} x+\ldots \ldots=\frac{1}{3}+\ldots \ldots\)

⇒ \(x^2-\frac{5}{3} x+\left(\frac{5}{6}\right)^2\)

⇒ \(\frac{1}{3}+\left(\frac{5}{6}\right)^2\)

⇒ \(\left[\text { adding }\left(\frac{\text { Coeff. of } x}{2}\right)^2 \text { on both sides }\right]\)

⇒ \(\left(x-\frac{5}{6}\right)^2=\frac{1}{3}+\frac{25}{36} \Rightarrow\left(x-\frac{5}{6}\right)^2=\frac{12+25}{36}\)

∴ \(\left(x-\frac{5}{6}\right)^2=\left(\frac{\sqrt{37}}{6}\right)^2\)

⇒ \(x-\frac{5}{6}= \pm \frac{\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} \text { and } \frac{5-\sqrt{37}}{6} \text {. }\)

Now, f(x) = \(3 x^2-5 x-1=3\left(x^2-\frac{5}{3} x\right)-1\)

⇒ \(3\left(x^2-\frac{5}{3} x+\frac{25}{36}-\frac{25}{36}\right)-1\)

⇒ \(3\left(x-\frac{5}{6}\right)^2-\frac{25}{12}-1=3\left(x-\frac{5}{6}\right)^2-\frac{37}{12}\)

which is of the form A(x- B)²– C, where A = 3, B = \(\frac{5}{6}\), C = \(\frac{37}{12}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Nature Of Roots Of A Quadratic Equation

The nature of roots of a quadratic equation ax² +bx + c = 0 depends on the value of its discriminant (D). i.e., upon b² – 4ac.

If a, b, and c are real numbers and a ≠ 0 then discriminant D = b²-4ac.

The value of discriminant affects the nature of roots in the following ways :

The roots of a quadratic equation are :

1. Real: When D > 0 i.e. (D > 0 or D = 0) (when quadratic equation can be expressed as in real linear factors, D > 0)
2. No Real (Imaginary): When D < 0
3. Real and Distinct: When D > 0
4. Real and Equal (Coincident): When D = 0

In this case each equal root will \(\left(\frac{-b}{2 a}\right)\)

Remember:

1.  ax-b>0 ⇒ \(x>\frac{b}{a} \text {, if } a>0 \text { and } x<\frac{b}{a} \text {, if } a<0\)
2. x²– a² > 0 ⇒ x<-a or x>a
3. x²– a² = 0 ⇒ x= —a or x =  a
4. x²– a² < 0 ⇒ x<a or x> -a ⇒ -a< x < 0
5. (x-a) (x-b)>0,a<b ⇒ x<a or x>b
6. (x- a) (x- b) < 0, a < b ⇒ a<x<b

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Nature Of Roots Of A Quadratic Equation Solved Examples

Question 1. Find the value of k so that the equation 2x² – 5x + k = 0 has two equal roots.

Solution:

The given equation is 2x² – 5x + K = 0

Comparing with ax² + bx + c = 0, we get

a = 2,b = -5 and c = k

The equation will have two equal roots if

D = 0

D = b²– 4ac = 0

or (-5)² -4×2×K = 0

25 – 8K = 0

⇒ \(k=\frac{25}{8}\)

The value of k =\(\frac{25}{8}\)

Question 2. The equation 3x² – 12x + (n – 5) = 0 has repeated roots. Find the value of n.

Solution:

Given equation is 3x²– 12x + (n – 5) = 0

Comparing with ax² +bx + c = 0, we get

a = 3, b = -12 and c = n- 5

The equation will have repeated (two equal) roots if

Discriminant (D) = 0

∴ D =b²-4ac=0

or (-12)²– 4 × 3 × (n- 5) = 0

⇒ 144 -12n + 60 = 0

⇒ 204- 12n = 0

⇒ -12n = -204

⇒ n = 17

Hence, the value of n is 17

Question 3. Find the value of k for which the equation 2 + k(2x +k- l) + 2 = 0 has real and equal roots.

Solution:

Given equation is

x² + k(2x + k – 1) + 2 = 0

x² + 2kx +k(k- 1) + 2 = 0

x² + 2kx + (k² -k + 2) = 0

Comparing with ax² + bx + c = 0,we get

a = 1, b = 2k and c = k² – k + 2

For real and equal roots,

Discriminant (D) = 0

∴ D =b²-4ac=0

or (2k)² – 4(1)(k² -k + 2)= 0

4k² – 4(k² -k + 2) = 0

k²– (k² -k + 2) = 0

K-2 = 0

k = 2.

Hence, the value of k is 2

Question 4. Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.

Solution:

Given equation is px² – 14x + 8 = 0

Let a and p be two roots of quadratic equation px²– 14x + 8 = 0, such that β = 6a

∴ Sum of roots = \(\frac{-b}{a}\)

⇒ \(\alpha+6 \alpha=-\frac{(-14)}{p}\)

⇒ \(7 \alpha=\frac{14}{p} \Rightarrow \alpha=\frac{2}{p}\)….(1)

Product of roots = \(\frac{c}{a}\)

⇒ \(\alpha \cdot 6 \alpha=\frac{8}{p}\)

⇒ \(6 \alpha^2=\frac{8}{p} \Rightarrow \alpha^2=\frac{8}{6 p}\) ….(2)

From equations (1 ) and (2), we get

⇒ \(\left(\frac{2}{p}\right)^2=\frac{8}{6 p} \Rightarrow \frac{4}{p^2}=\frac{4}{3 p}\)

P² = 3P

p² – 3p = 0

p(p-3) = 0 (don’t cancel p on both sides)

Either p = 0 or p = 3.

But p = 0 is not possible, as on putting, p = 0 in the given equation, we don’t have a quadratic equation and therefore we cannot get two roots.

Hence, P = 3

Alternatively,

Let one root of the quadratic equation px² – 14x + 8 = 0 is a.

∴ pα² – 14α + 8 =0 ….(1)

∴ Other root of the equation will be 6a.

p(6α)²– 14(6α) + 8 =0

36pα²– 84α + 8=0

9pα² – 21α + 2 =0 ….(2)

Solving equations (1) and (2) by cross-multiplication method.

⇒ \(\frac{\alpha^2}{-14(2)-8(-21)}=\frac{\alpha}{8(9 p)-2 p}=\frac{1}{p(-21)-9 p(-14)}\)

⇒ \(\frac{\alpha^2}{-28+168}=\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \( \frac{\alpha^2}{140}=\frac{1}{105 p}\text { and }\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \(\alpha^2=\frac{140}{105 p}=\frac{4}{3 p}\text { and }\alpha=\frac{70 p}{105 p}=\frac{2}{3}\)

⇒ \(\left(\frac{2}{3}\right)^2=\frac{4}{3 p} \quad \Rightarrow \quad \frac{4}{9}=\frac{4}{3 p} \quad \Rightarrow \quad 3 p=9\)

p = 3

The value of p = 3

Question 5. The equation x²+2(m-1)x+ (m + 5) 0 has real and equal roots. Find the value of m.

Solution:

Given equation is x²+ 2 (m – 1 )x + (m+ 5) = 0

Comparing with ax² + bx + c = 0, we get

a = 1, b = 2(m -1), c = m + 5

The equation will have two real and equal roots if

Discriminant (D) = 0

∴ D = b² – 4ac= 0

or [2(m-1)]2-4 × 1 × (m + 5) = 0

4(m² + 1 – 2m) – 4(m + 5) = 0

4m² + 4- 8m – 4m – 20 = 0

4m²– 12m, – 16 = 0

m²– 3m, -4 = 0

m² – 4m, + m, – 4 = 0

m(m, – 4) + 1 (m – 4) = 0

(m + 1) (m – 4) = 0

m = -1 or m = 4

Hence, the value(s) of m are -1 and 4.

Question 6. If -4 is a root of the equation x² + px- 4 = 0 and the equation x² + px + q = 0 has coincident roots, find the values of p and q.

Solution:

Since -4 is a root of x² + px- 4 = 0

Hence, (-4) will satisfy the equation.

Therefore, (-4)² +p(-4) -4=0

16 – 4p – 4 = 0

-4p +12 = 0

-4p = -12 .

p = 3 …….(1)

Given that, x² + px + q = 0 has coincident roots.

D = b² – 4ac = 0

D = p²-4×1×q=0

p²-4q = 0

3² -4q =0 [from (1)]

9 – 4q = 0

-4q = -9

q = \(\frac{9}{4}\)

Hence, the values of p = 3 and q = \(\frac{9}{4}\)

Question 7. Prove that both roots of the equation (x -a) (x- b) + (x- b) (x- c) + {x- c) (x-a) = 0 are real but they are equal only when a =b =c.

Solution:

The given equation may be written as

3x² – 2(a + b + c)x+ (ab + bc + ac) = 0

∴ Discriminant D = B² – 4AC

D = \(4(a+b+c)^2-4 \times 3(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2+2 a b+2 b c+2 a c\right)-12(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2-a b-b c-a c\right)\)

D = \(2\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 a c\right)\)

D = \(2\left[a^2+b^2-2 t b+b^2+c^2-2 b c+c^2+a^2-2 a c\right]\)

D = \(2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \geq 0\)

∵ \((a-b)^2 \geq 0,(b-c)^2 \geq 0 \text { and }(c-a)^2 \geq 0\)

Hence, both roots of the equation are real.

For equal root, we must have D = 0

⇒ (a – b)² + (b- c)² + (c- a)² = 0

⇒ a-b =0, b-c = 0, c-a = 0

⇒ a = b,b = c,c = a

⇒ a – b =c

Hence, roots are equal only when a=b = c

Question 8. Find the positive values of k for which the equations x² + kx + 64 = 0 and x² – 8x + k = 0 nail both have real roots:

Solution:

Given equations are

x² + kx + 64 = 0 …(1)

and x²-8x + k= 0 …(2)

Let D₁ and D₂ be discriminants of equations (1) and (2) respectively, then

D₁ = k² – 4 x 64 or  D₁ = k² – 656

and D₂ = (-8)² – 4k or D₂ = 64-4K

Both equations will have real roots, if

D₁ ≥ 0 and D₂ ≥ 0

⇒ K² -256 ≥ 0 and 64- 4k ≥ 0

⇒ k² ≥ 256 and64 ≥ 4k

⇒ k ≥ 16 and K ≤ 16

k = 16

Hence, both equations will have real roots, when k = 16.

The positive values of k = 16.

Question 9. Find the value(s) of k for which the given quadratic equations have real and distinct roots:

1. 2x² + kx + 4 = 0
2. 4x²-3kx+ 1=0
3. kx² + 6x + 1 = 0
4. x²-kx+ 9 = 0

Solution:

1. The given equation is 2x² + kx + 4 = 0

Comparing with ax² +bx + c = 0, we get

a-2, b=k and c = 4

∴ D = b²– 4ac ≥ 0 for real and distinct roots.

Therefore, D = k²– 4 × 2 × (4) ≥ 0

⇒ k² – 32 ≥ 0

⇒ k² ≥ 32

⇒ \(k \leq-4 \sqrt{2} \text { and } k \geq 4 \sqrt{2}\)

2. The given equation is 4x² – 3kx + 1 – 0

Comparing with ax² + bx + c = 0, wc get

a = 4, b = -3k and c = 1

∴ D = b² – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-3k)² – 4 × 4 × 1 ≥ 0

9k² -16>0 ⇒ 9k² > 16

⇒ \(k^2 \geq \frac{16}{9}\)

⇒ \(k \leq-\frac{4}{3} \quad \text { and } \quad k \geq \frac{4}{3}\)

3.  The given equation is kx² + 6x+1 = 0

Comparing with ax² +bx + c = 0, we get

a = k, b = 6 and c = 1

D = b² – 4ac > 0 for real and distinct roots

Therefore, D = (6)²– 4 × k × 1 ≥ 0

36-4K ≥ 0

36 ≥ 4K

k ≤ 9

4. The given equation is

x² -kx + 9 = 0

Comparing with ax² + bx + c = 0, we get

a = 1, b = -k and c = 9

∴ D = b² – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-k)²– 4 × 1 × 9 ≥ 0

⇒ K² – 36 ≥ 0

⇒ k ≤ -6 or K ≥ 6

Question 10. If roots of the equation (1 + m²)x² + 2mcx + (c² – a²) = 0 are equal, prove that : c² = a²(1+m²)

Solution:

We have,

(1 + m²)x² + 2mcx + (c² – a²) = 0

It has equal roots, if

D = 0

B²-4AC =0

⇒ (2mc)² – 4(1 + m²) (c² – a²) =0

⇒ 4m²c² – 4(c² – a² + m²c²– m²a²) = 0

⇒ m²c² -c² + a² – m²c² + m²a² = 0

⇒ c² = a²+ m²a² = a²(1 +m²)

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Word Problems Based On Quadratic Equations

To solve the word problem, first translate the words of the problem into an algebraic equation, then solve the resulting equation.

For solving a word problem based on a quadratic equation adopt the following steps:

Step 1: Read the statement of the problem carefully.

Step 2: Represent the unknown quantity of the problem by a variable.

Step 3: Translate the given statement to form an equation in terms of variables.

Step 4: Solve the equation.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equations Solved Examples

Question 1. The sum of a number and its reciprocal is \(\frac{10}{3}\), find the number(s).

Solution:

Let the number be x

∴ According to a given statement

⇒ \(x+\frac{1}{x}=\frac{10}{3}\)

⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)

⇒ \(3 x^2+3=10 x\)

⇒ \(3 x^2-10 x+3=0\)

⇒ \(3 x^2-(9+1) x+3=0\)

⇒ \(3 x^2-9 x-x+3=0\)

⇒ 3x(x-3)- l(x-3) = 0

⇒ (3x-1) (x-3) = 0

when 3x- 1 = 0  \(x=\frac{1}{3}\)

and when x-3 = 0,  x = 3

Hence, the number(s) are 3 and \(\frac{1}{3}\)

Question 2. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

Solution:

Let the larger part be x. Then, the smaller part = 16 – x

According to a given statement

⇒ 2x²-(16-x)² = 164

⇒ 2x² – (256 +x² – 32x) = 164

⇒ 2x² – 256- x² + 32x- 164 = 0

⇒ x² + 32x- 420 = 0

⇒ x² + 42x- 10x- 420 = 0

⇒ x(x + 42)- 10(x + 42) = 0

⇒ (x + 42) (x- 10) = 0

⇒ x = -42 or x =10

⇒ x = 10

Hence, the required parts are 10 and 6.

Question 3. The sum of squares of three consecutive natural numbers is 149, Find the numbers.

Solution:

Given

The sum of squares of three consecutive natural numbers is 149,

Let three consecutive natural numbers be a-, (x +1) and (x +2) respectively.

According to the given condition

⇒ \(x^2+(x+1)^2+(x+2)^2=149\)

⇒ \(x^2+\left(x^2+1+2 x\right)+\left(x^2+4+4 x\right)=149\)

⇒ \(3 x^2+6 x+5=149\)

⇒ \(3 x^2+6 x-144=0\)

⇒ \(x^2+2 x-48 =0\)

⇒ \(x^2+8 x-6 x-48=0\)

⇒ x(x+8)-6(x+8)=0

⇒ (x+8)(x-6)=0

⇒ x = -8 and x = 6

⇒ x = 6 (x = -8 is not a natural number)

Hence, the required natural numbers are 6, (6 + 1), (6 + 2) = 6, 7, 8 respectively.

Question 4. A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places. Find the number.

Solution:

Given

A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places.

Let the digit at the unit place be x and the digit at ten’s place be y

∴ Number = x × 1 + 10 × y

= x+ 10y.

After reversing the order of digits, the reversing number =y + 10x

According to the first condition,

xy = 8 ….(1)

According to the second condition,

⇒ (x+10y) – 63 -y + 10ar

⇒ 9y- 9x = 63

⇒ y -x = 7

⇒ y =x + 7 ….(2)

∴ from (1) and (2), we get

x (x + 7) = 8

⇒ x² +7x- 8 = 0

⇒ (x + 8) (x- 1) = 0

∴ x= 1 or x = -8

If \(\left.\begin{array}{l}
x=1 \\
y=1+7=8
\end{array}\right\} \quad \Rightarrow \text { number }=1+10(8)=81\)

If \(\left.\begin{array}{rl}
x=-8 \\
y=-8+7=-1
\end{array}\right\}\) not possible as digits cannot be negative.

So, required number = 81

Question 5. The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\), find the fraction.

Solution:

Given

The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\)

Let, the fraction be \(\frac{x}{y}\) where numerator is.v, then denominator = 2x + y

According to a given statement

⇒ \(\frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21}\)

Now, let \(\frac{x}{2 x+1}=a\)

⇒ \(a+\frac{1}{a}=\frac{58}{21} \Rightarrow \frac{a^2+1}{a}=\frac{58}{21}\)

⇒ \(21 a^2+21=58 a\)

⇒ \(21 a^2-58 a+21=0\)

⇒ \(21 a^2-49 a-9 a+21=0\)

⇒ \(7 a(3 a-7)-3(3 a-7)=0\)

⇒ \((7 a-3)(3 a-7)=0\)

⇒ \(a=\frac{3}{7} \quad \text { or } \quad a=\frac{7}{3}\)

when, \(a=\frac{3}{7} \Rightarrow \frac{x}{2 x+1}=\frac{3}{7}\)

⇒ 7x = 6x + 3

⇒ x = 3

when, \(a=\frac{7}{3} \Rightarrow \frac{x}{2 x+1}=\frac{7}{3}\)

⇒ 3x = 14x + 7

⇒ -11x = 7

⇒ \(\frac{-7}{11}\)

x = 3   \(\left(x=\frac{-7}{11} \text { is not a natural number }\right)\)

Hence, required fraction is \(\frac{x}{2 x+1}=\frac{3}{7} \text {.}\)

Question 6. The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse, find the sides of the triangle.

Solution:

Given

The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse

Let, the length of the shortest side be x meters

then hypotenuse = (2x + 6) metres

the third side = (2x + 6 – 2) metres

= (2x + 4) metres

Now, using Pythagoras theorem (2x+6)²=x² + (2x + 4)²

⇒ (4x² + 24x + 36)=x² + (4x²+ 16x + 16)

⇒ x² – 8x- 20 = 0

⇒ x² – 10x+2x- 20 = 0

⇒ x(x-10)+2(x+2) = 0

⇒ (x+10)(x+2) = 0

⇒ x = 10 or x = -2

⇒ x = 10

So, the length of the shortest side = 10 meters

length of the hypotenuse = (2 × 10 + 6) = 26 metres

and length of the third side = (2 × 10 + 4) = 24 metres

Hence, the sides of the triangle are 10 m, 24 m, and 26 m.

Question 7. The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son. Find their present ages.

Solution:

Given

The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son.

Let, the present age of the son be x years.

Hence, the age of father = 2×2

8 years hence, the age of son = (x + 8) years

and the age of father = (2x² + 8) years

According to a given statement

⇒ \(2 x^2+8=3(x+8)+4\)

⇒ \(2 x^2+8=3 x+24+4\)

⇒ \(2 x^2-3 x-20=0\)

⇒ \(2 x^2-8 x+5 x-20=0\)

⇒ \(2 x(x-4)+5(x-4)=0\)

⇒ (2 x+5)(x-4)=0

⇒ \(x=\frac{-5}{2}\) and x = 4

⇒ x = 4

Therefore, the present age of the son is 4 years and present age of the father is 2 × 4² = 32 years.

Question 8. Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank, then how much time will each tap take to fill the tank?

Solution:

Given

Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank

Let the time taken by TapI to fill a tank = x hrs

∴ The time taken by TapII to fill a tank = (x + 3) hrs

and time taken by both to fill a tank =  \(3 \frac{1}{13}=\frac{40}{13} \mathrm{hrs}\)

∴ Tap 1’s 1 hr work = \(\frac{1}{x}\)

Tap 2’s 1 hr work = \(\frac{1}{x+3}\)

and (Tap1 + Tap 2)’s 1hr work = \(\frac{13}{40}\)

⇒ \(\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}\)

⇒ \(\frac{x+3+x}{x(x+3)}=\frac{13}{40}\)

⇒ \(40(2 x+3)=13 x(x+3)\)

⇒ \(80 x+120=13 x^2+39 x\)

⇒ \(13 x^2-41 x-120=0\)

⇒ \(13 x^2-65 x+24 x-120=0\)

⇒ \(13 x(x-5)+24(x-5)=0\)

⇒ (x-5)(13 x+24)=0

∴ Either x-5 = 0 or 13x+ 24 = 0

x = 5 or \(x=\frac{-24}{13}\)

But time cannot be negative, so we reject x = \(\frac{-24}{13}\)

x = 5

Hence, time taken by Tap 1 = 5 hrs

and time taken by Tap 2 = (5 + 3) hrs = 8 hrs.

Question 9. A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes, find how much time B alone takes to complete the work.

Solution:

Given

A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes,

Let, B alone complete the work in* hours, then A alone will complete the work in (x- 6) hours.

⇒ \(\frac{1}{x-6}+\frac{1}{x}=\frac{3}{40}\)    \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\frac{x+x-6}{(x-6) x}=\frac{3}{40}\)

⇒ \(3 x^2-18 x=80 x-240\)

⇒ \(3 x^2-98 x+240=0\)

⇒ \(3 x^2-90 x-8 x+240=0\)

⇒ \(3 x(x-30)-8(x-30)=0\)

⇒ \((x-30)(3 x-8)=0\)

x = 30

⇒ \(x=\frac{8}{3}\)

x = 30   \(\left(\text { if } x=\frac{8}{3} \text {, then } x-6 \text { in negative }\right)\)

B alone will take 30 hours to complete the work.

Question 10. An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.

Solution:

Given

An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed.

Let the usual speed of an airplane be x km/hr. Given, distance = 1200 km

Time taken for journey of 1200 km = \(\frac{1200}{x} \text { hours }\)

⇒ \(\left(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\right)\)

When the speed is increased by 100 km/hr, time taken for the same journey = \(\frac{1200}{x+100} \text { hours. }\)

According to the given condition

⇒ \(\frac{1200}{x}-\frac{1200}{x+100}=1\)

⇒ \(\frac{1200(x+100)-1200 x}{x(x+100)}=1\)

⇒ \(1200 x+120000-1200 x=x^2+100 x\)

⇒ \(x^2+100 x-120000=0\)

⇒ \(x^2+400 x-300 x-120000=0\)

⇒ \(x(x+400)-300(x+400)=0\)

⇒ \((x+400)(x-300)=0\)

⇒ x = -400 or x = 300

⇒ x = 300 km/hr

∴ The usual speed = 300 km/hr.

Question 11. A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of water.

Solution:

Given

A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes.

Let, the speed of water be x km/hr.

Given the speed of a motor boat in still water is 15 km/hr.

Therefore, its speed downstream is (15 + x) km/hr and the speed upstream is (15- x) lon/hr.

Time taken for going 30 Ion downstream = \(\frac{30}{15+x} \text { hours. }\)

Time taken for going 30 Ion upstream = \(\frac{30}{15-x} \text { hours. }\)

According to the given condition

⇒ \(\frac{30}{15+x}+\frac{30}{15-x}=4+\frac{30}{60}\)

⇒ \(\frac{30(15-x)+30(15+x)}{(15+x)(15-x)}=\frac{9}{2}\)

⇒ \(\frac{450-30 x+450+30 x}{225-x^2}=\frac{9}{2}\)

⇒ \(900 \times 2=9\left(225-x^2\right)\)

⇒ \(1800=2025-9 x^2\)

⇒ \(9 x^2=225 \text { or } x^2=25\)

⇒ \(x= \pm 5\)

⇒ x=5

Hence, the speed of water is 5 Km/hr.

Question 12. A dealer sells an article for ₹24 and gains as much percent as the price of the article. Find the cost price of the article.

Solution:

Given

A dealer sells an article for ₹24 and gains as much percent as the price of the article.

Let, the C.P. of article be ₹x

Then, gain = x %

∴ \(\text { S.P. }=\frac{100+\text { gain } \%}{100} \times \text { C.P. } \quad\left(\text { or S.P. }=\text { C.P. }+ \text { C.P. } \times \frac{x}{100}\right)\)

⇒ \(24=\frac{100+x}{100} \times x\)

⇒ \(2400=100 x+x^2\)

⇒ \(x^2+100 x-2400=0\)

⇒ \(x^2+120 x-20 x-2400=0\)

⇒ \(x(x+120)-20(x+120)=0\)

⇒ \((x-20)(x+120)=0\)

⇒ x = 20 or x = -120

⇒ x = 20

Hence, the cost of the article is ₹20.

Question 13. A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter?

Solution:

Given

A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged.

Let, the length of the piece be A meters

Since the cost of A meters of cloth = ₹200

⇒ Cost of each metre of cloth = \(₹ \frac{200}{x}\)

New length of cloth = (x + 5)m

New cost of each metre of cloth = \(₹ \frac{200}{x+5}\)

Now, given \(\frac{200}{x}-\frac{200}{x+5}=2\)

⇒ \(\frac{200(x+5)-200 x}{x(x+5)}=2 \quad \Rightarrow \quad \frac{200 x+1000-200 x}{x^2+5 x}=2\)

⇒ \(1000=2 x^2+10 x\)

⇒ \(2 x^2+10 x-1000=0\)

⇒ \(x^2+5 x-500=0\)

⇒ \(x^2+25 x-20 x-500=0\)

⇒ \(x(x+25)-20(x+25)=0\)

⇒ \((x-20)(x+25)=0\)

x = 20 or x = -25

Since length cannot be negative

Hence, x = 20 m

and the original rate per metre = \(₹ \frac{200}{20}=₹ 10\)

Question 14. Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10. Find how many students went for a picnic.

Solution:

Given

Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10.

Let, no. of students who planned the picnic = x

Given, the budget for food = ₹480

∴ Share of each student = \(₹ \frac{480}{x}\)

Since eight of them failed to join the picnic

∴ No. of students went for picnic = (x- 8)

∴ Share of each student = \(₹ \frac{480}{x-8}\)

According to the given condition

⇒ \(\frac{480}{x-8}-\frac{480}{x}=10\)

⇒ \(\frac{480 x-480(x-8)}{(x-8) x}=10\)

⇒ \(\frac{480 x-480 x+3840}{x^2-8 x}=10\)

⇒ \(10 x^2-80 x=3840\)

⇒ \(x^2-8 x-384=0\)

⇒ \(x^2-24 x+16 x-384=0\)

⇒ \(x(x-24)+16(x-24)=0\)

⇒ (x- 24) (x + 16) =0 ⇒  x = 24    or  x = -16

Since no. of students cannot be negative

Hence, x = 24

No. of students who went for picnic = x- 8 = 24-8=16 students.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.1

Question 1. Check whether the following are quadratic equations:

1. \((x+1)^2=2(x-3)\)
2. \(x^2-2 x=(-2)(3-x)\)
3. (x-2)(x+1) =(x-1)(x+3)
4. \((x-3)(2 x+1)=x(x+5)\)
5. \((2 x-1)(x-3)=(x+5)(x-1)\)
6. \(x^2+3 x+1=(x-2)^2\)
7. \((x+2)^3=2 x\left(x^2-1\right)\)
8. \(x^3-4 x^2-x+1=(x-2)^3\)

Solution:

1. \((x+1)^2=2(x-3)\)

⇒ x² + 2x + 1 = 2x- 6

⇒ x² + 7 =0

The highest power of the variable x in it is 2.

∴ The given equation is a quadratic equation.

2. \(x^2-2 x=(-2)(3-x)\)

x² -2x =- 6 + 2x

x²-2x-2x+6 =0

x² -4x + 6 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

3. (x-2)(x+1) =(x-1)(x+3)

⇒ x(x + 1 ) -2 (x + 1 ) = x(x + 3) – 1 (x + 3)

⇒ x² +x-2x- 2 =x² + 3x-x-3

⇒ x² – x- 2 =x² + 2x- 3

⇒ x² +x-2- x² +3 = 0

⇒ -3x+1 = 0

The highest power of the variable x is not 2, in it.

∴ Given equation is not a quadratic equation.

4. (x-3) (2x+ 1) = x(x + 5)

x(2x+1)-3 (2x + 1) =x² + 5X

2x² + x- 6x- 3 – x² – 5x =0

x²– 10x – 3 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

5. (2x-1)(x-3)=(x +5)(x-1)

2x(x-3)-1(x-3) = x(x-1) + 5(x- 1)

2x² – 6x-x + 3 = x² -x+ 5x- 5

2x² – 7x + 3 = x² + 4x- 5

2x²– 7x + 3 -x² – 4x + 5 = 0

x²– 11x+ 8 = 0

The highest power of variable x is 2 in it.

∴ Given equation is a quadratic equation.

6. x² + 3x + 1 = (x- 2)²

x² + 3x + 1 =x² – 4x + 4

x² + 3x+ 1 -x² + 4x- 4 = 0

7x – 3 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is not a quadratic equation.

7. (x + 2)³ = 2x(x² – 1 )

⇒ x³ + 3.x.2 .(x + 2) + 2³ = 2x³ – 2x

⇒ x³+ 6x² + 12x + 8 – 2x³ + 2x =0

⇒ -x³+6x²+14x+8 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is a quadratic equation.

8. \(x^3-4 x^2-x+1=(x-2)^3\)

⇒ \(x^3-4 x^2-x+1=x^3-3 x \cdot 2(x-2)-2^3\)

⇒ \(x^3-4 x^2-x+1=x^3-6 x^2+12 x-8\)

⇒ \(x^3-4 x^2-x+1-x^3+6 x^2-12 x+8=0\)

⇒ \(2 x^2-13 x+9=0\)

Question 2. Represent the following situations in the form of quadratic equations :

1. The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of tire plot.
2. The product of two consecutive positive integers is 306. We need to find the integers.
3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
4. A train travels a distance of 480 1cm at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:

1. Let the breadth of the plot =x meter

∴ Length of plot = (2x + 1) metre

Now, from length x breadth = area

(2x + 1) × x =528

⇒ 2x² + x =528

⇒ 2x² + x- 528 =0

which is the required quadratic equation.

2. Let two consecutive positive integers are x and x+ 1

∵ The product of two consecutive positive integers = 306

∴ x(x+1) = 360

⇒ x²+x = 306

⇒ x²+x-306 = 0

which is the required quadratic equation.

3. Let the present age of Rohan = x years

∴ Present age of Rohan’s mother

= (x + 26), years

After 3 years,

Rohan’s age = (x + 3) years

The age of Rohan’s mother = (x + 26 + 3) years

= (x + 29) years

According to the problem,

After 3 years, the product of their ages = 360

(x + 3) (x + 29) = 360

x(x + 29) + 3(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 – 360 = 0

x² + 32x- 273 = 0

which is the required quadratic equation.

4. Let the speed of train = A km/hr

Distance = 480 1cm

∴ Time taken to cover a distance of 480 km

⇒ \(\frac{480}{x} \mathrm{hrs}\)

If, the speed of the train = (x- 8) Km/hr  then the time is taken to cover 480 Km distance

⇒ \(\frac{480}{x-8} \mathrm{hrs}\)

According to the problem,

⇒ \(\frac{480}{x-8}-\frac{480}{x}=3\)

⇒ \(\frac{480 x-480(x-8)}{x(x-8)}=3\)

⇒ 480x- 480x + 3840 = 3x(x- 8)

⇒ 3840 = 3(x² – 8x)

⇒ 1280 =x²– 8x

⇒ 0= x² -8x -1280

⇒ x² -8x- 1280 = 0

which is the required quadratic equation.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.2

Question 1. Find the roots of the following quadratic equations by factorisation:

1. x²-3x-10 = 0
2. 2x²+x-6 = 0
3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0\)
4. \(2 x^2-x+\frac{1}{8}=0\)
5. 100x²– 20x + 1=0

Solution:

1. x²-3x-10 = 0

⇒ x²-5 x+2 x-10=0

⇒ x(x-5)+2(x-5)=0

⇒ (x-5)(x+2)=0

⇒ 5=0 or x+2=0

⇒ x=5 or x=-2

∴ Roots of given quadratic equation = 5, -2

2. 2 x^2+x-6=0

⇒ 2 x^2+4 x-3 x-6=0

⇒ 2 x(x+2)-3(x+2)=0

⇒ (x+2)(2 x-3)=0

⇒ x+2 =0 or 2x-3 = 0

⇒ x = -2 or 2x = 3

⇒ x =-2 or x = \(\frac{3}{2}\)

∴ Roots of given quadratic equation = -2, \(\frac{3}{2}\)

3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \)

⇒ \(\sqrt{2} x^2+5 x+2 x+5 \sqrt{2}=0\)

⇒ \((\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5)=0\)

⇒ \((\sqrt{2} x+5)(x+\sqrt{2})=0\)

⇒ \(\sqrt{2} x+5=0 \quad \text { or } x+\sqrt{2}=0\)

⇒ \(\sqrt{2} x=-5 \text { or }x=-\sqrt{2}\)

⇒ \(x=\frac{-5}{\sqrt{2}} \text { or }x=-\sqrt{2}\)

∴ Roots of given quadratic equation

⇒ \(\frac{-5}{\sqrt{2}},-\sqrt{2} \text {. }\)

4. \(2 x^2-x+\frac{1}{8}=0\)

⇒ \(16 x^2-8 x+1=0\)

⇒ \(16 x^2-4 x-4 x+1=0\)

⇒ \(4 x(4 x-1)-1(4 x-1)=0\)

⇒ (4x-1)(4 x-1)=0

⇒ 4x-1 = 0  4x- 1 = 0

⇒ 4x= 1 or 4x= 1

⇒ \(x=\frac{1}{4} \quad \text { or } \quad x=\frac{1}{4}\)

∴ Roots of given quadratic equation = \(\frac{1}{4}, \frac{1}{4} \text {.}\)

5. \(100 x^2-20 x+1=0\)

⇒ \(100 x^2-10 x-10 x+1=0\)

⇒ 10x(10 x-1)-1(10 x-1)=0

⇒ (10x-1)(10 x-1)=0

⇒ 10x-1=0 or 10 x-1=0

⇒ 10x=1 or 10 x=1

⇒ \(x=\frac{1}{10} \text { or } \quad x=\frac{1}{10}\)

∴ Roots of given quadratic equation \(\frac{1}{10}, \frac{1}{10}\)

Question 2. Represent the following situations mathematically:

1. John and Jiwanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
2. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.

Solution : 

1. Let the number of marbles initially with John =x

∴ Initially, the number of marbles with Jiwanti = 45-x

After losing 5 marbles each,

Remaining marbles with John =x- 5

Remaining marbles with Jiwanti = 45 -x- 5

= 40-x

According to the problem,

Product of remaining marbles with them =124

(x- 5) (40-x) = 124

⇒ x(40-x) -5(40-x) = 124

⇒ 40x-x²– 200 + 5x= 124

⇒ -x² + 45x- 200 = 124

⇒ 0= 124 +x²– 45x + 200

⇒ x² – 45x + 324 = 0

⇒ x²– 36x -9x + 324 = 0

⇒ x(x- 36) -9(x- 36) = 0

⇒ (x-36) (x-9) = 0

⇒ x-36 = 0 or x-9 = 0

⇒ x = 36 or x = 9

If x = 36 then 45-x = 45-36 = 9

If x = 9 then 45-x = 45-9 = 36

∴ Marbles with John = 36

and marbles with Jiwanti = 9

or

Marbles with John = 9

and marbles with Jiwanti = 36

2. Let the number of toys = x

Cost of each toy =  ₹(55 -x)

Cost of x toys = ₹(55 – x) x-

₹(55x-x2)

According to the problem,

⇒ 55x -x² = 750

⇒ 0 =x²-55X+750

⇒ x² – 30x- 25x + 750 =0

⇒ x(x- 30) -25 (x-30) =0

⇒ (x-30) (x-25) =0

⇒ x-30 = 0 or x- 25 = 0

⇒ x = 30 or x = 25

∴ Number of toys produced = 25 or 30.

Question 3. Find two numbers whose sum is 27 and whose product is 182.

Solution:

Let one number = x

∴ Second number =27 -x

According to the problem,

Product of two numbers = 182

⇒ x (27-x) = 182

⇒ 27x -x² = 182

⇒ 0 =x² – 27x+ 182

⇒ x²– 13x- 14x+ 182 =0

⇒ x(x- 13) -14 (x- 13) =0

⇒ (x- 13) (x- 14) =0

⇒ x-13 = 0 or x-14 =0

⇒ x= 13 or x = 14

If x = 13, then 27-x = 27-13 = 14

If x = 14, then 27 -14 = 13

Therefore, numbers=(13 and 14)or(14and 13).

Question 4. Find two consecutive positive integers, a sum of whose squares is 365.

Solution:

Given

The sum of whose squares is 365

Let two consecutive positive integers be x and x + 1.

According to the problem,

x² + (x+1)² =365

x² + x² + 2x + 1 =365

2x² + 2x + 1 – 365 = 0

2x² + 2x- 364 = 0

x²+x- 182 =0

x²+14x-13x-182 = 0

=x (x + 14) (x- 13) =0

x + 14=0 or x-13=0

x = -14 or x = 13

x is a positive integer,

∴ neglecting x = -14,

x = 13

x+1 = 13+1 = 14

Therefore, required positive integers = 13 and 14.

Question 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Given

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm

Let the base of the right triangle = x cm

∴ Its height = (x- 7) cm

Given, the hypotenuse of the right triangle = 13cm

From Pythagoras theorem, in a right triangle (base)² + (height)² = (hypotenuse)²

⇒ x² + (x- 7)² = 132

⇒ x²+x²– 14x + 49 = 169

⇒ 2x² – 14x + 49 — 169 =0

⇒ 2x² -14x- 120 =0

⇒ x² – 7x- 60 =0

⇒ x²– 12x + 5x -60 =0

⇒ x(x- 12) + 5(x- 12) =0

⇒ (x- 12) (x+ 5) =0

⇒ x-12 =0 or x + 5= 0

⇒ x = 12 or x = -5

but the value of x cannot be negative.

∴ Neglecting

x = -5

x = 12

⇒ x-7 =12-7 = 5

Therefore, the other two sides of the triangle =12 cm and 5 cm.

Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.

Solution:

Given

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90,

Let the number of pottery articles produced in a day = x

∴ Cost of each article = ₹(2x+ 3)

⇒ Cost of x articles = ₹(2x + 3)x

According to the problem,

(2x + 3) x = 90

⇒ 2x² + 3x = 90

⇒ 2x² + 3x-90 =0

⇒ 2x² + 15x- 12x- 90 = 0

⇒ x(2x+ 15) -6(2x+ 15) =0

⇒ (2x+ 1 5) (x -6) =0

⇒ 2x + 15 = 0 or x -6 = 0

⇒ \(x=-\frac{15}{2} \text { or } \quad x=6\)

but x = \(-\frac{15}{2}\) is not possible

∴ x = 6

⇒ 2x + 3 = 2×6 + 3=15

Therefore, the number of pottery articles in a day = 6, and the cost of each article =₹15.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.3

Question 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :

1. 2x² – 7x + 3 = 0
2. 2x² + x- 4 = 0
3. \(4 x^2+4 \sqrt{3} x+3=0\)
4. 2x² +x + 4 = 0

Solution:

1. 2x² – 7x + 3 = 0

⇒ \(x^2-\frac{7}{2} x+\frac{3}{2}=0\)

⇒ \( {\left[x^2-2 x \cdot \frac{7}{4}+\left(\frac{7}{4}\right)^2\right]+\frac{3}{2}-\left(\frac{7}{4}\right)^2 }=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2+\frac{24-49}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\frac{25}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\left(\frac{5}{4}\right)^2=0\)

⇒ \(\left(x-\frac{7}{4}-\frac{5}{4}\right)\left(x-\frac{7}{4}+\frac{5}{4}\right)=0\)

⇒ \((x-3)\left(x-\frac{1}{2}\right)=0\)

⇒ \(x-3=0\text { or }x-\frac{1}{2}=0\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of given equation = \(3, \frac{1}{2}\)

2. \(2 x^2+x-4=0\)

⇒ \(x^2+\frac{1}{2} x-2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2-2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{32+1}{16}\right)=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\frac{33}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{\sqrt{33}}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}-\frac{\sqrt{33}}{4}\right)\left(x+\frac{1}{4}+\frac{\sqrt{33}}{4}\right)=0\)

⇒ \(x+\frac{1}{4}-\frac{\sqrt{33}}{4}=0 \text { or } x+\frac{1}{4}+\frac{\sqrt{33}}{4}=0\)

⇒ \(x=\frac{\sqrt{33}}{4}-\frac{1}{4} \quad \text { or } \quad x=-\frac{\sqrt{33}}{4}-\frac{1}{4}\)

⇒ \(x=\frac{\sqrt{33}-1}{4} \quad \text { or } \quad x=-\left(\frac{\sqrt{33}+1}{4}\right)\)

∴ Roots of a given equation

⇒ \(\frac{\sqrt{33}-1}{4},-\left(\frac{\sqrt{33}+1}{4}\right)\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

⇒ \(x^2+\sqrt{3} x+\frac{3}{4}=0\)

⇒ \(x^2+2 x \cdot \frac{\sqrt{3}}{2}+\left(\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(\left(x+\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(x+\frac{\sqrt{3}}{2}=0 \text { or }x+\frac{\sqrt{3}}{2}=0\)

⇒ \(x=-\frac{\sqrt{3}}{2}\text { or }x=-\frac{\sqrt{3}}{2}\)

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. \(2 x^2+x+4=0\)

⇒ \(x^2+\frac{1}{2} x+2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2+2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2+\frac{32-1}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2=\frac{-31}{16}\)

⇒ \(x+\frac{1}{4}=\sqrt{-\frac{31}{16}}\) which is an imaginary number.

∴ Roots of given equation does not exist.

Question 2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Solution:

1. 2x² – 7x + 3 = 0

On comparing with ax² +bx + c,

a = 2,b=-7,c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-7) \pm \sqrt{(-7)^2-4(2)(3)}}{2(2)}\)

⇒ \(\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm \sqrt{25}}{4}=\frac{7 \pm 5}{4}\)

⇒ \(x=\frac{7+5}{4}\text { or }x=\frac{7-5}{4}\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of equation = \(3, \frac{1}{2}\)

2. 2x² +x- 4 = 0

On comparing with ax² + bx + c = 0

a = 2,b = 1,c = -4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(-4)}}{2 \times 2}=\frac{-1 \pm \sqrt{1+32}}{4}\)

⇒ \(\frac{-1 \pm \sqrt{33}}{4}\)

⇒ \(x=\frac{-1+\sqrt{33}}{4} \text { or } \frac{-1-\sqrt{33}}{4}\)

∴ Roots of equation = \(\frac{-1+\sqrt{33}}{4}, \frac{-1-\sqrt{33}}{4}\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

On comparing with ax² +bx + c = 0

a = 4, b = \(4 \sqrt{3}\), c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-4 \sqrt{3} \pm \sqrt{(4 \sqrt{3})^2-4 \times 4 \times 3}}{2 \times 4}\)

⇒ \(\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}=\frac{-4 \sqrt{3} \pm \sqrt{0}}{8}\)

⇒ \(-\frac{4 \sqrt{3}}{8}=-\frac{\sqrt{3}}{2}\)

Number of roots of a quadratic equation = 2

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. 2x² + x + 4 = 0

On comparing with ax², + bx + c = 0

a = 2,b=1,c = 4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(4)}}{2 \times 2}\)

⇒ \(\frac{-1 \pm \sqrt{1-32}}{4}=\frac{-1 \pm \sqrt{-31}}{4}\)

⇒ \(\sqrt{-31}\)  is an imaginary number,

∴ the values for are imaginary.

So, the real roots of the given equation do not exist.

Question 3. Find the roots of the following equations:

1. \(x-\frac{1}{x}=3, x \neq 0\)
2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

Solution:

1. \(x-\frac{1}{x}=3, x \neq 0\)

⇒ \(\frac{x^2-1}{x}=3\)

⇒ \(x^2-1=3 x \Rightarrow x^2-3 x-1=0\)

On comparing with ax² + bx + c = 0

a= 1, b = -3, c =-1

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)}=\frac{3 \pm \sqrt{9+4}}{2}\)

⇒ \(\frac{3 \pm \sqrt{13}}{2}\)

⇒ \(x=\frac{3+\sqrt{13}}{2} \text { or } \quad x=\frac{3-\sqrt{13}}{2}\)

Therefore, the roots of given equations

⇒ \(\frac{3+\sqrt{13}}{2}, \frac{3-\sqrt{13}}{2} \text {. }\)

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

⇒ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)

⇒ \(\frac{x-7-x-4}{x(x-7)+4(x-7)}=\frac{11}{30}\)

⇒ \(\frac{-11}{x^2-7 x+4 x-28}=\frac{11}{30}\)

⇒ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)

⇒ \(x^2-3 x-28=-30\)

⇒ \(x^2-3 x-28+30=0\)

⇒ \(x^2-3 x+2=0\)

⇒ \(x^2-2 x-x+2=0\)

⇒ x(x-2)-1(x-2)=0

⇒ (x-2)(x-1)=0

⇒ x-2 = 0 or x-1 = 0

⇒ x = 2 or x = 1

∴ Roots of given equation = 2,1

Question 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.

Solution:

Given

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\).

Let the present age of Rehman = x years

3 years before, Rehman’s age = (x- 3) years

After 5 years, Rehman’s age = (x + 5) years

According to the problem, \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

⇒ \(\frac{(x+5)+(x-3)}{(x-3)(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x(x+5)-3(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+5 x-3 x-15}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+2 x-15}=\frac{1}{3}\)

⇒ \(x^2+2 x-15=6 x+6\)

⇒ \(x^2+2 x-15-6 x-6=0\)

⇒ \(x^2-4 x-21=0\)

⇒ \(x^2-7 x+3 x-21=0\)

⇒ x(x-7)+3(x-7)=0

⇒ (x-7)(x+3)=0

⇒ x-7=0 or  x+3=0

⇒ x=7 or  x=-3

but the age cannot be negative.

∴ x = 7

⇒ Rehman’s age = 7 years.

Question 5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solution:

Given

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210.

Let marks obtained by Shefali in Mathematics =x

∴ Her marks in English = (30- x)

If, marks in Mathematics = x + 2

marks in English =30-x-3 = 27-x

then (x + 2) (27 -x) =210

⇒ x(27 -x) + 2(27 -x) =210

⇒ 27x -x² + 54 – 2x = 210

⇒ 25x-x² + 54 =210

⇒ 0 = x²– 25x- 54 + 210

⇒ x² – 25x + 156 =0

⇒ x²– 13x- 12x+ 156 =0

⇒ x(x -13) – 12(x – 13) =0

⇒ (x- 13) (x- 12) =0

⇒ x -13 = 0 or x-12=0

⇒ x -13 or x-12

If x = 13 then 30-x = 30-13 = 17

If x = 12 then 30-x = 30-12 = 18

So, for Shefali

Maries in Mathematics = 13

and marks in English = 17

or

Maries in Mathematics = 12

and marks in English = 18

Question 6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution:

Given

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side,

Let the smaller side of the rectangular field =xmetre = breadth

Larger side = (x + 30) metre = length

diagonal = (x + 60) metre

Now, (length) + (breadth) = (diagonal)

⇒ x² + (x+ 30)² = (x+60)²

⇒ x² +x² + 60x + 900 =x² + 120x + 3600

⇒ x² +x² + 60x + 900 -x² – 120x- 3600 = 0

⇒ x²– 60x -2700 = 0

⇒ x²– 90x + 3x -2700 =0

⇒ x(x- 90) + 30(x- 90) = 0

⇒ (x-90) (x + 30) =0

⇒ x-90 =0 or x+30 = 0

⇒ x = 90 or x = -30

but the side cannot be negative.

⇒ x = 90

⇒ x+30 = 90+30 = 120

Therefore, sides of the rectangular field = 120 m and 90 m.

Question 7. The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Given

The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number.

Let smaller number =x

∴ Larger number × 8 = x²

⇒ Larger number = \(\frac{x^2}{8}\)

According to the problem, (larger number)² – (smaller number)² =180

⇒ \(\left(\frac{x^2}{8}\right)^2-x^2=180\)

⇒ \(\left(\frac{y}{8}\right)^2-y=180\)

⇒ \(\text { where } x^2=y \text { (say) }\)

⇒ \(\frac{y^2}{64}-y=180\)

⇒ \(y^2-64 y=11\)

⇒ \(y^2-64 y-11520=0\)

⇒ \(y^2-144 y+80 y-11520=0\)

⇒ \(y(y-144)+80(y-144)=0\)

⇒ \((y-144)(y+80)=0\)

⇒ \(y-144=0 \text { or } y+80=0\)

⇒ \(y=144 \text { or } y=-80\)

⇒ \(x^2=144 \text { or } x^2=-80\)

x² =- 80 is not possible.

∴ \(x^2=144\)

⇒ \(x= \pm 12\)

⇒ \(x=12 \quad \text { or } \quad x=-12\)

⇒ \(\frac{x^2}{8}=\frac{144}{8}=18\)

Therefore, a number are 12, 18 or -12, 18.

Question 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Given

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey.

Let the speed of the train =x km/hr

∴ Time taken to cover 360 km distance = \(\frac{360}{x} \mathrm{~km}\)

If the speed of train = (x+ 5) km/hr

then time taken to cover 360 km distance = \(\frac{360}{x+5} \mathrm{hr}\)

According to the problem, \(\frac{360}{x}-\frac{360}{x+5}=1\)

⇒ \(\frac{360(x+5)-360 x}{x(x+5)}=1\)

⇒ \(\frac{360 x+1800-360 x}{x^2+5 x}=1\)

⇒ \(\frac{1800}{x^2+5 x}=1\)

⇒ \(x^2+5 x=1800\)

⇒ \(x^2+5 x-1800=0\)

⇒ \(x^2+45 x-40 x-1800=0\)

⇒ x(x+45)-40(x+45)=0

⇒ (x+45)(x-40)=0

⇒ x+45=0  or x-40=0

⇒ x=-45  or  x=40

but speed cannot be negative.

∴ Speed of train = 40 km/hr.

Question 9. Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution:

Given

Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately.

Let the time taken to fill the tank by the tap of smaller diameter =x hours

Time taken to fill the tank by the tap of larger diameter = (x- 10) hours

Now, work done by tap of smaller diameter in 1 hour = \(\frac{1}{x}\)

and work done by tap of larger diameter in 1hour = \(\frac{1}{x}+\frac{1}{x-10}\)

Work done by two taps in 1 hour = \(\frac{1}{x}+\frac{1}{x-10}\)

According to the problem, both taps fill the Time taken by the passenger train to cover 132 1cm tank in \(9 \frac{3}{8}=\frac{75}{8} \)

∴ \(\left(\frac{1}{x}+\frac{1}{x-10}\right) \times \frac{75}{8}=1\)

⇒ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\)

⇒ \(\frac{x-10+x}{x(x-10)}=\frac{8}{75}\)

⇒ \(\frac{2 x-10}{x^2-10 x}=\frac{8}{75}\)

⇒ \(8 x^2-80 x=150 x-750\)

⇒ \(8 x^2-80 x-150 x+750=0\)

⇒ \(8 x^2-230 x+750=0\)

⇒ \(4 x^2-115 x+375=0\)

⇒ \(4 x^2-100 x-15 x+375=0\)

⇒ 4 x(x-25)-15(x-25)=0

⇒ (x-25)(4 x-15)=0

⇒ x-25=0  or  4 x-15=0

⇒ x-25 or \(x=\frac{15}{4}\)

but \(x=\frac{15}{4}\) is not possible because both taps fill the tank in \(9 \frac{3}{4}\) hours.

⇒ x = 25 and x- 10 = 25 – 10= 15

Therefore, the tap with a smaller diameter fills the tank in 25 hours and the tap with a larger diameter fills the tank in 15 hours.

Question 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Solution:

Given

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train

Let the average speed of passenger train =x km/hr.

∴ Average speed of express train = (x+11) km/hr

Time taken by passenger train to cover 1321cm distance = \(\frac{132}{x} \mathrm{hr}\)

and time taken by express train to cover 1321cm distance = \(\frac{132}{x+11} \mathrm{hr}\)

According to the problem,

⇒ \(\frac{132}{x}-\frac{132}{x+11}=1\)

⇒ \(\frac{132(x+11)-132 x}{x(x+11)}=1\)

⇒ \(\frac{132 x+1452-132 x}{x^2+11 x}=1\)

⇒ \(x^2+11 x=1452\)

⇒ \(x^2+11 x-1452=0\)

⇒ \(x^2+44 x-33 x-1452=0\)

⇒ x(x+44)-33(x+44)=0

⇒ (x+44)(x-33)=0

⇒ x+44=0  or  x-33=0

⇒ x=-44  or  x=33

but the speed cannot be negative.

∴ x = 33

⇒ x + 11 =33+ 11 =44

Average speed of passenger train = 33 km/hr

an average speed of express train = 44 km/hr

Question 11. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:

Given

The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m

Let side of a square = x metre

Perimeter of this square = 4x metre

∴ Perimeter of second square = (4x + 24) metre

⇒ Side of second square = \(\frac{4 x+24}{4}\) = (x + 6) metre

Now, area of first square =x2 m2 and area of second square = (x+ 6)2 m2

According to the problem,

⇒ \((x+6)^2+x^2=468\)

⇒ \(x^2+12 x+36+x^2-468=0\)

⇒ \(2 x^2+12 x-432=0\)

⇒ \(x^2+6 x-216=0\)

⇒ \(x^2+18 x-12 x-216=0\)

⇒ x(x+18)-12(x+18)=0

⇒ (x+18)(x-12)=0

⇒ x+18=0   or   x-12=0

⇒ x=-18   or x=12

but the side of a square cannot be negative.

∴ x= 12

⇒ x+ 6 = 12 + 6= 18

Therefore, sides of squares = 12 m and 18 m

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Exercise 4.4

Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

1. \(2 x^2-3 x+5=0\)
2. \(3 x^2-4 \sqrt{3} x+4=0\)
3. \(2 x^2-6 x+3=0\)

Solution:

1. \(2 x^2-3 x+5=0\)

On comparing with ax² + bx + c = 0

a = 2, b=-3,c = 5

Discriminant D = b² – 4ac = (-3)² – 4(2) (5)

= 9-40 =-31

∵ D is negative

∴ Roots of the equation are imaginary.

Here, the real roots of the equation do not exist.

2. \(3 x^2-4 \sqrt{3} x+4=0\)

On comparing with ax² + bx + c = 0

a = 3,b =\(-4 \sqrt{3}\), c= 4

Discriminant D = b²– 4ac

⇒ \((-4 \sqrt{3})^2-4(3)(4)\)

= 48-48 = 0

The roots of given equation are real and equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \sqrt{3} \pm 0}{2 \times 3}=\frac{2}{\sqrt{3}}\)

Roots of given equation = \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. 2x² – 6x + 3=0

On comparing with ax² + bx + c = 0

a = 2, b = – 6, c = 3

∴ Discriminant D = b²– 4ac = (-6)²– 4 (2) (3)

= 36-24 = 12 > 0

∵ D is positive

Roots of the given equation are real and unequal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{6 \pm \sqrt{12}}{2 \times 2}=\frac{6 \pm 2 \sqrt{3}}{4}\)

⇒ \(x=\frac{3 \pm \sqrt{3}}{2}\)

⇒ \(\frac{3+\sqrt{3}}{2} \text { or } \frac{3-\sqrt{3}}{2} \text {. }\)

Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

1. 2x²+kx+ 3 = 0
2. k x(x- 2) + 6 = 0

Solution:

2x²+kx+ 3 = 0

On comparing with ax² +bx + c = 0

a = 2, b=k, c = 3

DiscriminantD = b²– 4ac = k² – 4 x 2 x 3

= k² -24

Given that, the roots are equal.

∴ D = 0  ⇒ k² – 24 = 0

⇒ k² = 24

⇒ K = \(\pm \sqrt{24}= \pm 2 \sqrt{6}\)

2.  kx(x -2) + 6 = 0

kx² – 2kx+6 = 0

On comparing with a² + bx + c = 0

a = k, b = -2k, c = 6

∴ Discriminant D = b²– 4ac = (-2k)² – 4k(6)

= 4k²– 24k = 4k(k- 6)

Given that the roots are equal.

∴ D = 0

⇒ 4k(k – 6) = 0

⇒ k – 6 = 0

⇒ k = 6

Question 3. Is It possible to design a rectangular mango grove whose length Is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution:

Let the breadth of mango grove

= x metre

∴ Length = 2x metre

Area of grove = length x breadth

= (2x)(x) = 2x²

According to the problem,

⇒ 2x² = 800

⇒ x² = 400

⇒ x = ± 20

but the breadth cannot be negative.

∴ x = 20

2x = 2 × 20 = 40

Therefore, a grove is possible, and its length = 40m and breadth = 20 m.

Question 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.

Solution:

Let the present age of one friend =  x years

∴ Present age of second friend = (20- x) years

4 years ago, the age of first friend = (x- 4) years

The age of second friend = 20 – x- 4

= (16- x) years

According to the problem,

(x-4) (16 -x) =48

⇒ x(1 6 – 4) -4(16 – x) =48

⇒ 16x – x²– 64 + 4x = 48

⇒ 20x-x² -64 = 48

⇒ 0 = 48 – 20x + x² + 64

x²-20x+ 112=0 ….(1)

On comparing with ax² + bx + c = 0

a = 1, b = -20, c = 112

Now, discriminant D=b² – 4ac

= (- 20)² – 4 (1) (112)

= 400 -448 =-48 <0

Roots of equation (1) are imaginary.

Therefore, the given condition is not possible.

Question 5. Is it possible to design a rectangular park of perimeter 80 m and an area of 400 m2? If so, find its length and breadth.

Solution:

Let the length of the park = x meter

Now, 2 (length + breadth) = perimeter

2(x + breadth) = 80 metre

x+ breadth = 40 metre

breadth = (40 – x) metre

According to the problem,

Area of park= 400 m²

x(40 – x) = 400

40x-x² = 400

0 = x² – 40x + 400

x² – 40x + 400 = 0

On comparing with ax² +bx+ c = 0

a = 1, b = -40, c = 400

Discriminant D = b² – 4ac

= (-40)² – 4(1)(400)

= 1600- 1600 = 0

⇒ Roots are equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{40 \pm 0}{2 \times 1}=20\)

So, such a park is possible.

Its length = breadth = 20m.

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation Multiple Choice Questions

Question 1. Which of die following equations has the root 3?

1. \(x^2+x+1\)
2. \(x^2-4 x+3=0\)
3. \(3 x^2+x-1=0\)
4. \(x^2+9=0\)

Answer: 2. \(x^2-4 x+3=0\)

Question 2. The sum of roots of the equation 5x² – 3x + 2 = 0 is

1. \(\frac{3}{5}\)
2. \(-\frac{3}{5}\)
3. \(\frac{2}{5}\)
4. \(-\frac{2}{5}\)

Answer: 1. \(\frac{3}{5}\)

Question 3. The quadratic equation \(2 x^2-\sqrt{5} x+1=0\) has

1. Two distinct real roots
2. Two equal roots
3. No real root
4. More than two real roots

Answer: 3. No real root

Question 4. One root of the equation x² + k x + 4 = 0 is -2. The value of k is :

1. -2
2. 2
3. -4
4. 4

Answer: 4. 4

Question 5. The value of k for which die roots of equation 2kx² – 6x + 1 = 0 is equal, is :

1. \(-\frac{9}{2}\)
2. \(\frac{9}{2}\)
3. 9
4. -9

Answer: 2. \(\frac{9}{2}\)

Question 6. Which of the following is a quadratic equation?

1. (x + 2)² =x²– 5x + 3
2. x³ +x² = (x- 1)³
3. 3x² + 1 = (3x- 2) (x + 5)
4. 5x- 7 = 1 + x

Answer: 2. x³ +x² = (x- 1)³

Question 7. If the product of roots of the equation 5×2- 3x + /{ = 0 is 2, then the value of k is:

1. 1
2. 2
3. 5
4. 10

Answer: 4. 10

Question 8. The discriminant of quadratic equadon 3xx³ +x² = (x- 1)³– 6x + 4 = 0 is :

1. 12
2. 13
3. -12
4. \(3 \sqrt{6}\)

Answer: 3. -12

Question 9. The roots of the quadratic equation x²– 4 = 0 are :

1. ± 0.2
2. ±1
3. ± 2
4. ±4

Answer: 3. ± 2

Question 10. The discriminant of equation \(3 x^2-2 x+\frac{1}{3}=0\) will be:

1. 3
2. 2
3. 1
4. 0

Answer: 4. 0

Question 11. If the roots of the quadratic equation 3x² – 12x + m = 0 are equal, then the value of m will be :

1. 4
2. 7
3. 9
4. 12

Answer: 4. 12