Class 8 Maths Solution | Lesson 12 Quadratic Equations by Factorization Method | CDC

Solving quadratic equations by factorization method

Exercise 12.2

1. Solve :

(a) x² - 3x = 0

Solution:

Here,

x² - 3x = 0

or, x(x - 3) = 0

Either, x = 0

∴ x = 0

or (x - 3) = 0

∴ x = 3

Hence, the values of x are 0 and 3.

(b) 2x² - x = 0

Solution:

Here,

2x² - x = 0

or, x(2x - 1) = 0

Either, x = 0

∴ x = 0

or (2x - 1) = 0

∴ x = 1/2

Hence, the values of x are 0 and 1/2.

(c) 9x² + 3x = 0

Solution:

Here,

9x² + 3x = 0

or, x(9x + 3) = 0

Either, x = 0

∴ x = 0

or (9x + 3) = 0

∴ x = -1/3

Hence, the values of x are 0 and -1/3.

(d) 9y² - 4 = 0

Solution:

Here,

9y² - 4 = 0

or, (3y - 2)(3y + 2) = 0

Either, (3y - 2) = 0

∴ y = 2/3

or (3y + 2) = 0

∴ y = -2/3

Hence, the values of y are 2/3 and -2/3.

(e) 5x + 9x² = 0

Solution:

Here,

5x + 9x² = 0

or, x(5 + 9x) = 0

Either, x = 0

∴ x = 0

or (5 + 9x) = 0

∴ x = -5/9

Hence, the values of x are 0 and -5/9.

(f) 4y² - 7y = 0

Solution:

Here,

4y² - 7y = 0

or, y(4y - 7) = 0

Either, y = 0

∴ y = 0

or (4y - 7) = 0

∴ y = 7/4

Hence, the values of y are 0 and 7/4.

(g) x² - 49 = 0

Solution:

Here,

x² - 49 = 0

or, (x - 7)(x + 7) = 0

Either, (x - 7) = 0

∴ x = 7

or (x + 7) = 0

∴ x = -7

Hence, the values of x are 7 and -7.

(h) 169x² - 96 = 0

Solution:

Here,

169x² - 96 = 0

or, x² = 96/169

∴ x = ±√(96/169)

∴ x = ±(4√6)/13

Hence, the values of x are (4√6)/13 and -(4√6)/13.

(i) x²/4 - 36 = 0

Solution:

Here,

x²/4 - 36 = 0

or, x²/4 = 36

∴ x² = 144

∴ x = ±12

Hence, the values of x are 12 and -12.

(j) 5x² - 125 = 0

Solution:

Here,

5x² - 125 = 0

or, 5(x² - 25) = 0

∴ (x² - 25) = 0

or, (x - 5)(x + 5) = 0

Either, (x - 5) = 0

∴ x = 5

or (x + 5) = 0

∴ x = -5

Hence, the values of x are 5 and -5.

(k) x² - 7 = 29

Solution:

Here,

x² - 7 = 29

or, x² = 29 + 7

∴ x² = 36

∴ x = ±6

Hence, the values of x are 6 and -6.

(l) x² - 4x = 0

Solution:

Here,

x² - 4x = 0

or, x(x - 4) = 0

Either, x = 0

∴ x = 0

or (x - 4) = 0

∴ x = 4

Hence, the values of x are 0 and 4.

2. Solve : (a) x² - 3x = 0

Solution:

Here,

x² - 3x = 0

or, x (x - 3) = 0

Either, x = 0

∴ x = 0

or (x - 3) = 0

∴ x = 3

Hence, the values of x are 0 and 3.

(b) 2x² - x = 0

Solution:

Here,

2x² - x = 0

or, x(2x - 1) = 0

Either, x = 0

∴ x = 0

or (2x - 1) = 0

∴ x = 1/2

Hence, the values of x are 0 and 1/2.

(c) 9x² + 3x = 0

Solution:

Here,

9x² + 3x = 0

or, x (9x + 3) = 0

Either, x = 0

∴ x = 0

or (9x + 3) = 0

∴ x = -1/3

Hence, the values of x are 0 and -1/3.

(d) 9y² - 4 = 0

Solution:

Here,

9y² - 4 = 0

or, (3y - 2)(3y + 2) = 0

Either, (3y - 2) = 0

∴ y = 2/3

or (3y + 2) = 0

∴ y = -2/3

Hence, the values of y are 2/3 and -2/3.

(e) 5x + 9x² = 0

Solution:

Here,

5x + 9x² = 0

or, x(5 + 9x) = 0

Either, x = 0

∴ x = 0

or (5 + 9x) = 0

∴ x = -5/9

Hence, the values of x are 0 and -5/9.

(f) 4y² - 7y = 0

Solution:

Here,

4y² - 7y = 0

or, y(4y - 7) = 0

Either, y = 0

∴ y = 0

or (4y - 7) = 0

∴ y = 7/4

Hence, the values of y are 0 and 7/4.

(g) x² - 49 = 0

Solution:

Here,

x² - 49 = 0

or, (x - 7)(x + 7) = 0

Either, (x - 7) = 0

∴ x = 7

or (x + 7) = 0

∴ x = -7

Hence, the values of x are 7 and -7.

(h) 169x² - 96 = 0

Solution:

Here,

169x² - 96 = 0

or, x² = 96/169

∴ x = ±√(96/169)

∴ x = ±(4√6)/13

Hence, the values of x are (4√6)/13 and -(4√6)/13.

(i) x²/4 - 36 = 0

Solution:

Here,

x²/4 - 36 = 0

or, x²/4 = 36

∴ x² = 144

∴ x = ±12

Hence, the values of x are 12 and -12.

(j) 5x² - 125 = 0

Solution:

Here,

5x² - 125 = 0

or, 5(x² - 25) = 0

∴ (x² - 25) = 0

or, (x - 5)(x + 5) = 0

Either, (x - 5) = 0

∴ x = 5

or (x + 5) = 0

∴ x = -5

Hence, the values of x are 5 and -5.

(k) x² - 7 = 29

Solution:

Here,

x² - 7 = 29

or, x² = 29 + 7

∴ x² = 36

∴ x = ±6

Hence, the values of x are 6 and -6.

(l) x² - 4x = 0

Solution:

Here,

x² - 4x = 0

or, x(x - 4) = 0

Either, x = 0

∴ x = 0

or (x - 4) = 0

∴ x = 4

Hence, the values of x are 0 and 4.

(m) x² - 6x + 8 = 0

Solution:

Here,

x² - 6x + 8 = 0

or, x² - 4x - 2x + 8 = 0

or, x(x - 4) - 2(x - 4) = 0

or, (x - 4)(x - 2) = 0

Either, (x - 4) = 0

∴ x = 4

or (x - 2) = 0

∴ x = 2

Hence, the values of x are 4 and 2.

(n) 2x² - x - 6 = 0

Solution:

Here,

2x² - x - 6 = 0

or, 2x² - 4x + 3x - 6 = 0

or, 2x(x - 2) + 3(x - 2) = 0

or, (2x + 3)(x - 2) = 0

Either, (2x + 3) = 0

∴ x = -3/2

or (x - 2) = 0

∴ x = 2

Hence, the values of x are -3/2 and 2.

(o) y² + 7y + 12 = 0

Solution:

Here,

y² + 7y + 12 = 0

or, y² + 4y + 3y + 12 = 0

or, y(y + 4) + 3(y + 4) = 0

or, (y + 4)(y + 3) = 0

Either, (y + 4) = 0

∴ y = -4

or (y + 3) = 0

∴ y = -3

Hence, the values of y are -4 and -3.

(p) 7x² + 13x - 2 = 0

Solution:

Here,

7x² + 13x - 2 = 0

or, 7x² + 14x - x - 2 = 0

or, 7x(x + 2) - 1(x + 2) = 0

or, (7x - 1)(x + 2) = 0

Either, (7x - 1) = 0

∴ x = 1/7

or (x + 2) = 0

∴ x = -2

Hence, the values of x are 1/7 and -2.

(q) x² + 9x - 22 = 0

Solution:

Here,

x² + 9x - 22 = 0

or, x² + 11x - 2x - 22 = 0

or, x(x + 11) - 2(x + 11) = 0

or, (x + 11)(x - 2) = 0

Either, (x + 11) = 0

∴ x = -11

or (x - 2) = 0

∴ x = 2

Hence, the values of x are -11 and 2.

(r) x² - 18x + 77 = 0

Solution:

Here,

x² - 18x + 77 = 0

or, x² - 11x - 7x + 77 = 0

or, x(x - 11) - 7(x - 11) = 0

or, (x - 11)(x - 7) = 0

Either, (x - 11) = 0

∴ x = 11

or (x - 7) = 0

∴ x = 7

Hence, the values of x are 11 and 7.

(s) 2x² + 11x + 12 = 0

Solution:

Here,

2x² + 11x + 12 = 0

or, 2x² + 8x + 3x + 12 = 0

or, 2x(x + 4) + 3(x + 4) = 0

or, (2x + 3)(x + 4) = 0

Either, (2x + 3) = 0

∴ x = -3/2

or (x + 4) = 0

∴ x = -4

Hence, the values of x are -3/2 and -4.

(t) 3x² - 11x - 20 = 0

Solution:

Here,

3x² - 11x - 20 = 0

or, 3x² - 15x + 4x - 20 = 0

or, 3x(x - 5) + 4(x - 5) = 0

or, (3x + 4)(x - 5) = 0

Either, (3x + 4) = 0

∴ x = -4/3

or (x - 5) = 0

∴ x = 5

Hence, the values of x are -4/3 and 5.

(u) 10x² + 19x + 6 = 0

Solution:

Here,

10x² + 19x + 6 = 0

or, 10x² + 15x + 4x + 6 = 0

or, 5x(2x + 3) + 2(2x + 3) = 0

or, (5x + 2)(2x + 3) = 0

Either, (5x + 2) = 0

∴ x = -2/5

or (2x + 3) = 0

∴ x = -3/2

Hence, the values of x are -2/5 and -3/2.

(v) 12x² - 11x + 2 = 0

Solution:

Here,

12x² - 11x + 2 = 0

or, 12x² - 8x - 3x + 2 = 0

or, 4x(3x - 2) - 1(3x - 2) = 0

or, (4x - 1)(3x - 2) = 0

Either, (4x - 1) = 0

∴ x = 1/4

or (3x - 2) = 0

∴ x = 2/3

Hence, the values of x are 1/4 and 2/3.

(w) 3z² - 11z + 6 = 0

Solution:

Here,

3z² - 11z + 6 = 0

or, 3z² - 9z - 2z + 6 = 0

or, 3z(z - 3) - 2(z - 3) = 0

or, (3z - 2)(z - 3) = 0

Either, (3z - 2) = 0

∴ z = 2/3

or (z - 3) = 0

∴ z = 3

Hence, the values of z are 2/3 and 3.

(x) (x + 1)² - 4 = 0

Solution:

Here,

(x + 1)² - 4 = 0

or, (x + 1 - 2)(x + 1 + 2) = 0

or, (x - 1)(x + 3) = 0

Either, (x - 1) = 0

∴ x = 1

or (x + 3) = 0

∴ x = -3

Hence, the values of x are 1 and -3.

(y) (p + 3)² - 16 = 0

Solution:

Here,

(p + 3)² - 16 = 0

or, (p + 3 - 4)(p + 3 + 4) = 0

or, (p - 1)(p + 7) = 0

Either, (p - 1) = 0

∴ p = 1

or (p + 7) = 0

∴ p = -7

Hence, the values of p are 1 and -7.

(z) (x + 6)² - 36 = 0

Solution:

Here,

(x + 6)² - 36 = 0

or, (x + 6 - 6)(x + 6 + 6) = 0

or, x(x + 12) = 0

Either, x = 0

or (x + 12) = 0

∴ x = -12

Hence, the values of x are 0 and -12.

(aa) (x - 7)² - 64 = 0

Solution:

Here,

(x - 7)² - 64 = 0

or, (x - 7 - 8)(x - 7 + 8) = 0

or, (x - 15)(x + 1) = 0

Either, (x - 15) = 0

∴ x = 15

or (x + 1) = 0

∴ x = -1

Hence, the values of x are 15 and -1.

(ab) 100 - (x - 5)² = 0

Solution:

Here,

100 - (x - 5)² = 0

or, (10 + x - 5)(10 - x + 5) = 0

or, (x + 5)(15 - x) = 0

Either, (x + 5) = 0

∴ x = -5

or (15 - x) = 0

∴ x = 15

Hence, the values of x are 15 and -5.

(3a) Find the quadratic equation whose roots (value of x) are 1 and 2.

Solution:

The standard form of a quadratic equation is:

x² - (sum of roots)x + (product of roots) = 0

Here, roots are 1 and 2.

Sum of roots = 1 + 2 = 3

Product of roots = 1 × 2 = 2

Therefore, the quadratic equation is:

x² - 3x + 2 = 0

(4b) Find the quadratic equation whose roots (value of x) are 3 and -2.

Solution:

The standard form of a quadratic equation is:

x² - (sum of roots)x + (product of roots) = 0

Here, roots are 3 and -2.

Sum of roots = 3 + (-2) = 1

Product of roots = 3 × (-2) = -6

Therefore, the quadratic equation is:

x² - 1x - 6 = 0

or, x² - x - 6 = 0

(5c) Find the quadratic equation whose roots (value of y) are 3 and -2.

Solution:

The standard form of a quadratic equation is:

y² - (sum of roots)y + (product of roots) = 0

Here, roots are 3 and -2.

Sum of roots = 3 + (-2) = 1

Product of roots = 3 × (-2) = -6

Therefore, the quadratic equation is:

y² - 1y - 6 = 0

or, y² - y - 6 = 0