## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 7 Quadratic Equations Chapter Test

Solve the following (1 to 3) equations :

Question 1.

(i) x(2x + 5) = 3

(ii) 3x^{2} – 4x – 4 = 0

Solution:

(i) x(2x + 5) = 3

{∵ 2 ×(-3) = -6

∴ – 6 = 6 × (-1)

5 = 6 – 1}

⇒ 2x^{2} + 5x – 3 = 0

⇒ 2x^{2} + 6x – x – 3 = 0

⇒ 2x(x + 3) – 1(x + 3) = 0

⇒ (x + 3)(2x – 1) = 0

Either, x + 3 = 0, then x = -3

or 2x – 1 =0, then 2x = 1 ⇒ x =

∴ x = -3,

(ii) 3x^{2} – 4x – 4 = 0

⇒ 3x^{2} – 6x + 2x – 4 = 0

{∵ 3 × (-4) = -12

∴ -12 = -6 × 2

-4 = -6 + 2}

⇒ 3x(x – 2) + 2(x – 2) = 0

⇒ (x – 2) (3x + 2) = 0

Either, x – 2 = 0, then x = 2

or 3x + 2 = 0, then 3x = -2 ⇒ x =

∴ x = 2,

Question 2.

(i) 4x^{2} – 2x + = 0

(ii) 2x^{2} + 7x + 6 = 0

Solution:

(i) 4x^{2} – 2x + = 0

⇒ 16x^{2} – 8x + 1 = 0

⇒ 16x^{2} – 8x + 1 =0

⇒ 16x^{2} – 4x – 4x + 1 =0

⇒ 4x(4x – 1) – 1 (4x – 1) = 0

⇒ (4x – 1) (4x – 1) = 0 ⇒ (4x – 1)^{2} = 0

⇒ 4x – 1 = 0 => 4x = 1

∴ x = ,

(ii) 2x^{2} + 7x + 6 = 0

{∵ 2 × 6 = 12

∴ 12 = 3 × 4

7 = 3 + 4}

⇒ 2x(x + 2) + 3(x + 2) = 0

⇒ (x + 2) (2x + 3) = 0

Either, x + 2 = 0. then x = -2

or 2x + 3 = 0, then 2x = -3 ⇒ x =

∴ x = -2,

Question 3.

Solution:

⇒ 10x^{2} – 60x + 80 = 6x^{2} – 30x + 30

⇒ 10x^{2} – 60x + 80 – 6x^{2} + 30x – 30 = 0

⇒ 4x^{2} – 30x + 50 = 0

⇒ 2x^{2} – 15x + 25 = 0

⇒ 2x^{2} – 10x – 5x + 25 = 0

{∵ 2 × 25 = 50

∴ 50 = -10 × (-5)

-15 = -10 – 5}

⇒ 2x(x – 5) – 5(x – 5) = 0

⇒ (x – 5) (2x – 5) = 0

Either, x – 5 = 0, then x = 5

or 2x – 5 = 0, then 2x = 5 ⇒ x =

∴ x = 5,

⇒ 4x^{2} – 8x – 6x + 12 = x^{2} – x

⇒ 4x^{2} – 14x + 12 – x^{2} + x = 0

⇒ 3x^{2} – 13x + 12 = 0

⇒ 3x^{2} – 4x – 9x + 12 = 0

{∵ 3 × 12 = 36

∴ 36 = (-4) × (-9)

-13 = -4 – 9}

⇒ x(3x – 4) – 3(3x – 4) = 0

⇒ (3x – 4) (x – 3) = 0

Either, 3x – 4 = 0, then 3x = 4 ⇒ x =

or x – 3 = 0, then x = 3

∴ x = 3,

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