# R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4A

R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations exercise 4A are provided here. This exercise helps students in solving quadratic equations by factorisation. Subject experts have crafted all the solutions using images and graphs for clear understanding. Download free R S Aggarwal Solutions of Class 10 pdf for Maths and score well in the exams.

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### Access other exercise solutions of Class 10 Maths Chapter 4 Quadratic Equations

Exercise 4B Solutions: 16 Questions (Short Answers)

Exercise 4C Solutions: 15 Questions (Short Answers)

Exercise 4D Solutions: 20 Questions (Long Answers)

## Exercise 4A Page No: 175

Question 1: Which of the following are quadratic equations in x?

1. x2 â€“ x + 3 = 0
2. 2x2 + Â 5/2 x – âˆš3 = 0
3. âˆš2x2 + 7x + 5âˆš2 = 0
4. 1/3x2 + 1/5x â€“ 2 = 0
5. x2 – 3x – âˆšx + 4 = 0
6. x â€“ 6/x = 3
7. x + 2/x = x2
8. x^2 â€“ 1/x^2 = 5
9. (x + 2)^3 = x^3 – 8
10. (2x + 3)(3x + 2) = 6(x â€“ 1)(x â€“ 2)
11. (x + 1/x)^2 = 2)x + 1/x) + 3

Solution:

AÂ quadratic equationÂ is anÂ equationÂ of the second degree.

(i) x2 â€“ x + 3 = 0

Highest Degree: 2

(ii) 2x2 +Â 5/2x – âˆš3 = 0

Highest Degree: 2

(iii) âˆš2x2+7x+5âˆš2=0

Highest Degree: 2

(iv) 1/3x2 + 1/5x â€“ 2 = 0

Above equation can be simplify as: 5x2 + 3x â€“ 2 = 0

Highest Degree: 2

(v) x2 – 3x – âˆšx + 4 = 0

Equation has a fractional power.

(vi) x â€“ 6/x = 3

Simply as x2 – 3x – 6 = 0

Degree : 2

(vii) x + 2/x = x2

Simplify above equation:

x3 – x2 – 2 = 0

Degree: 3

(viii) x^2 â€“ 1/x^2 = 5

Simplify above equation

x4 Â  – 1 = 5 x2

or x4-5x2-1 =0

Degree: 4

(ix) (x + 2)^3 = x^3 â€“ 8

x^3 + 8 + 6x^2 + 12x = x^3 â€“ 8

-6x^2 + 12x + 16 = 0

Degree = 2

(x) (2x + 3)(3x + 2) = 6(x â€“ 1)(x â€“ 2)

Simplify above equation:

6x^2 + 4x + 9x + 6 = 6x^2 â€“ 12x â€“ 6x + 12

31x â€“ 6 = 0

Degree : 1

(xi) (x + 1/x)^2 = 2)x + 1/x) + 3

Simplify above equation:

(x^4 + 2x^2 + 1) / x^2 = (2x^2 + 2) / x + 3

(x^4 + 2x^2 + 1)x = x^2(2x^2 + 2) + 3

Answer: (i), (ii), (iii), (iv), (vi) and (ix) are only quadratic equations.

Question 2:

Which of the following are the roots of 3xÂ² + 2x â€“ 1=0?

(i) -1

(ii) 1/3

(iii) -1/2

Solution:

Simplify given equation:

3xÂ² + 2x â€“ 1 = 3xÂ² + 3x â€“ x â€“ 1

= 3x (x + 1) â€“ 1 (x + 1)

= (x + 1) (3x â€“ 1)

To find roots, put 3xÂ² + 2x â€“ 1 = 0

Either, x + 1 = 0 or 3x â€“ 1 =0

x = -1 or x = 1/3

Therefore, (-1) and 1/3 are the required roots.

Question 3:

(i) Find the value of k for which x = 1 is a root of the equation xÂ² + kx + 3 = 0. Also, find the other root.

(ii) Find the values of a and 6 for which x = 3/4 and x = -2 are the roots of the equation axÂ² + bx â€“ 6 = 0.

Solution:

(i) x = 1 is a solution of x^2+kx+3=0, which means it must satisfy the equation.

(1)^2 + k(1) + 3 = 0

k = -4

Hence the required value of k = -4

Find other root:

We have equation, x^2 – 4x + 3 = 0

x^2 – x – 3x + 3 = 0

x(x – 1) – 3(x – 1) = 0

(x – 1)(x – 3) = 0

either x – 1 = 0 or x – 3 = 3

x = 1 or x = 3

Other root is 3.

(ii) given equation is axÂ² + bx â€“ 6 = 0

As Â¾ is its root, then must satisfy the equation

a(3/4)Â² + b(3/4) â€“ 6 = 0

9a +12b â€“ 96 = 0 â€¦.(1)

Again, x = -2 is its root

a(-2)Â² + b(-2) â€“ 6 = 0

4a â€“ 2b â€“ 6 = 0 â€¦(2)

Solving (1) and (2), we get

a = 4 and b = 5

Question 4: Show that x = -bc/ad is a solution of the quadratic equation

Solve each of the following quadratic equations.

Question 5:

(2x â€“ 3)(3x + 1) = 0

Solution:

(2x â€“ 3)(3x + 1) = 0

Either 2x â€“ 3 = 0 or 3x + 1 = 0

x = 3/2 or x = -1/3

Question 6: 4xÂ² + 5x = 0

Solution:

4x2 + 5x = 0

Or x (4x + 5) = 0

Either x = 0 or 4x + 5 = 0, then

x = -5/4 or 0

Question 7: 3xÂ² â€“ 243 = 0

Solution:

3xÂ² â€“ 243 = 0

or xÂ² â€“ 81 = 0

(x)Â² â€“ (9)Â² = 0

(x + 9) (x â€“ 9) = 0

Either, x + 9 = 0 or x â€“ 9 = 0

x = -9 or 9

Question 8: 2xÂ² + x â€“ 6 = 0

Solution:

2×2 ï€« x ï€­ 6 ï€½ 0

2×2 ï€« 4x ï€­ 3x ï€­ 6 ï€½ 0

Either x ï€« 2 ï€½ 0 or 2x ï€­ 3 ï€½ 0

x = -2 or 3/2

Question 9: xÂ² + 6x + 5 = 0

Solution:

x^2 + 6x + 5 = 0

x^2 + x + 5x + 5 = 0

x(x + 1) + 5(x + 1) = 0

(x + 5)(x + 1) = 0

either x +5 = 0 or x + 1 = 0

x = -5 or -1

Question 10: 9xÂ² â€“ 3x â€“ 2 = 0

Solution:

9x^2 â€“ 3x â€“ 2 = 0

9x^2 â€“ 6x + 3x â€“ 2 = 0

3x(3x -2) + (3x â€“ 2) = 0

(3x +1)(3x â€“ 2) = 0

either (3x +1) = 0 or (3x â€“ 2) = 0

x = -1/3 or 2/3

Question 11: xÂ² + 12x + 35 = 0

Solution:

xÂ² + 12x + 35 = 0

xÂ² + 7x + 5x + 35 = 0

x(x + 7) + 5((x + 7) = 0

(x + 5)(x + 7) = 0

either (x + 5) = 0 or (x + 7) = 0

x = -5 or -7

Question 12: xÂ² = 18x â€“ 77

Solution:

x^2 – 18x + 77 = 0

x^2 – 7x – 11x + 77 = 0

x (x – 7) – 11(x – 7) = 0

(x – 11)(x – 7) = 0

either (x – 11) = 0 or (x – 7) = 0

x = 11 or 7

Question 13: 6xÂ² + 11x + 3 = 0

Solution:

6x^2 + 11x + 3 = 0

6x^2 + 2x + 9x + 3 = 0

2x(3x + 1) + 3(3x + 1) = 0

(2x + 3)(3x + 1) = 0

either (2x + 3) = o or (3x + 1) = 0

x = -1/3 or -3/2

Question 14: 6xÂ² + x â€“ 12 = 0

Solution:

Question 15:

3xÂ² â€“ 2x â€“ 1 = 0

Solution:

Question 16:

4xÂ² â€“ 9x = 100

Solution:

4xÂ² â€“ 9x = 100

4xÂ² â€“ 9x â€“ 100 = 0

Question 17:

15xÂ² â€“ 28 = x

Solution:

15x2 ï€­ 28 ï€½ x

15x2 ï€­ x ï€­ 28 ï€½ 0

15x2 ï€­ ï€¨21x ï€­ 20xï€©ï€­ 28 ï€½ 0

15x2 ï€­ 21x ï€« 20x ï€­ 28 ï€½ 0

3x ï€« 4 ï€½ 0 or 5x ï€­ 7 ï€½ 0

x = -4/3 or 7/5

Question 18:

4 â€“ 11x = 3xÂ²

Solution:

4 ï€­11x ï€½ 3x2

3x2 ï€«11x ï€­ 4 ï€½ 0

3x2 ï€«12x ï€­ x ï€­ 4 ï€½ 0

Either x ï€« 4 ï€½ 0 or 3x ï€­1 ï€½ 0

x = -4 or 1/3

Question 19: 48xÂ² â€“ 13x â€“ 1 = 0

Solution:

48x2 ï€­13x ï€­1 ï€½ 0

48x2 ï€­ï€¨16x ï€­ 3xï€©ï€­1 ï€½ 0

48x2 ï€­16x ï€« 3x ï€­1 ï€½ 0

Either 16x ï€«1 ï€½ 0 or 3x ï€­1 ï€½ 0

x = -1/16 or 1/3

Question 20:

xÂ² + 2âˆš2 x â€“ 6 = 0

Solution:

Question 21;

âˆš3 xÂ² + 10x + 7âˆš3 = 0

Solution:

âˆš3 xÂ² + 10x + 7âˆš3 = 0

âˆš3 xÂ² + 10x + 7âˆš3 = 0

âˆš3 xÂ² + 3x + 7x + 7âˆš3 = 0

âˆš3 x(x + âˆš3) + 7( x + âˆš3) = 0

(x + âˆš3)(âˆš3 x + 7) = 0

either âˆš3 x + 7 = 0 or x + âˆš3 = 0

x = -âˆš3 or -7/âˆš3

Question 22:

âˆš3 xÂ² + 11x + 6âˆš3 = 0

Solution:

âˆš3 xÂ² + 11x + 6âˆš3 = 0

âˆš3 xÂ² + 9x + 2x + 6âˆš3 = 0

âˆš3 x (x + 3âˆš3) + 2(x + 3âˆš3) = 0

(âˆš3x + 2)(x + 3âˆš3) = 0

either (âˆš3x + 2) = 0 or (x + 3âˆš3) = 0

x = -3âˆš3 or -2âˆš3/3

Question 23:

3âˆš7 xÂ² + 4x + âˆš7 = 0

Solution:

3âˆš7 xÂ² + 4x – âˆš7 = 0

3âˆš7 xÂ² + 4x – âˆš7 = 0

3âˆš7 xÂ² -3x + 7x – âˆš7 = 0

3x(âˆš7x – 1) + âˆš7(âˆš7x – 1) = 0

(3x + âˆš7)(âˆš7x – 1) = 0

either (3x + âˆš7) = 0 or (âˆš7x – 1) = 0

x = -âˆš7/3 or 1/âˆš7

Question 24:

âˆš7 xÂ² â€“ 6x â€“ 13âˆš7 = 0

Solution:

âˆš7 xÂ² â€“ 6x â€“ 13âˆš7 = 0

Question 25: 4âˆš6 xÂ² â€“ 13x â€“ 2âˆš6 = 0

Solution:

Question 26: 3xÂ² â€“ 2âˆš6x + 2 = 0

Solution:

3xÂ² â€“ 2âˆš6x + 2 = 0

Question 27: âˆš3 xÂ² â€“ 2âˆš2 x â€“ 2âˆš3 = 0

Solution:

âˆš3 xÂ² â€“ 2âˆš2 x â€“ 2âˆš3 = 0

âˆš3 xÂ² â€“ 3âˆš2 x + âˆš2 xâ€“ 2âˆš3 = 0

âˆš3x (x – âˆš6) + âˆš2(x – âˆš6) = 0

(âˆš3x+ âˆš2)(x – âˆš6) = 0

either (âˆš3x+ âˆš2) = 0 or (x – âˆš6) = 0

x = âˆš6 or -âˆš2/âˆš3

Question 28: xÂ² â€“ 3âˆš5 x + 10 = 0

Solution:

Question 29:

xÂ² â€“ (âˆš3 + 1) x + âˆš3 = 0

Solution:

xÂ² â€“ (âˆš3 + 1) x + âˆš3 = 0

xÂ² â€“ (âˆš3 + 1) x + âˆš3 = 0

xÂ² â€“ âˆš3x – x + âˆš3 = 0

x(x – âˆš3) – (x – âˆš3) = 0

(x – 1)(x – âˆš3) = 0

either (x – 1) = 0 or (x – âˆš3) = 0

x = 1 or âˆš3

Question 30:

xÂ² + 3âˆš3 x â€“ 30 = 0

Solution:

xÂ² + 3âˆš3 x â€“ 30 = 0

xÂ² + 5âˆš3 x – 2 âˆš3 â€“ 30 = 0

x(x + 5âˆš3) – 2âˆš3 (x + 5âˆš3) = 0

(x – 2âˆš3)(x + 5âˆš3) = 0

either (x – 2âˆš3) = 0 or (x + 5âˆš3) = 0

x = -5âˆš3 or 2 âˆš3

## R S Aggarwal Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4A

Class 10 Maths Chapter 4 Quadratic Equations Exercise 4A is based on the topics: