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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.4

Question 1.

Solve:

(i) (x – 5) (x – 7) (x + 6) (x + 4) = 504

(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16

Solution:

(i) (x – 5) (x + 4) (x – 7) (x + 6) = 504

(x^{2} – x – 20) (x^{2} – x – 42) = 504

Let y = (x^{2} – x)

(y – 20) (y – 42) = 504

⇒ y^{2} – 42y – 20y + 840 = 504

⇒ y^{2} – 62y + 336 = 0

⇒ (y – 56) (y – 6) = 0

⇒ (y – 56) = 0 or (y – 6) = 0

⇒ x^{2} – x – 56 = 0 or x^{2} – x – 6 = 0

⇒ (x – 8) (x + 7) = 0 or (x – 3) (x + 2) = 0

⇒ x = 8, -7 or x = 3, -2

The roots are 8, -7, 3, -2

(ii) (x – 4) (x – 7) (x – 2) (x + 1) = 16

⇒ (x – 4) (x – 2) (x – 7) (x + 1) = 16

⇒ (x^{2} – 6x + 8) (x^{2} – 6x – 7) = 16

Let x^{2} – 6x = y

(y + 8)(y – 7) = 16

⇒ y^{2} – 7y + 8y – 56 – 16 = 0

⇒ y^{2} + y – 72 = 0

⇒ (y + 9) (y – 8) = 0

y + 9 = 0

x^{2} – 6x + 9 = 0

(x – 3)^{2} = 0

x = 3, 3

or

y – 8 = 0

x^{2} – 6x – 8 = 0

The roots are 3, 3, 3 ±√17

Question 2.

Solve : (2x – 1) (x + 3) (x – 2) (2x + 3) + 20 = 0.

Solution:

(2x – 1) (2x + 3) (x + 3) (x – 2) + 20 = 0

⇒ (4x^{2} + 6x – 2x – 3) (x^{2} – 2x + 3x – 6) + 20 = 0

⇒ (4x^{2} + 4x – 3) (x^{2} + x – 6) + 20 = 0

⇒ [4(x^{2} + x) – 3] [x^{2} + x – 6] + 20 = 0

Let y = x^{2} + x

⇒ (4y – 3) (y – 6) + 20 = 0

⇒ 4y^{2} – 24y – 3y + 18 + 20 = 0

⇒ 4y^{2} – 27y + 38 = 0

⇒ (4y – 19) (y – 2) = 0

(4y – 19) = 0

4(x^{2} + x) – 19 = 0

4x^{2} + 4x – 19 = 0

or

(y – 2) = 0

x^{2} + x – 2 = 0

(x + 2) (x – 1) = 0

x = -2, +1

The roots are -2, 1, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

### Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.4 Additional Questions

Questions 1.

Solve (x – 3) (x – 6) (x – 1) (x + 2) + 54 = 0.

Solution:

(x – 3) (x – 1) (x – 6) (x + 2) + 54 = 0

(x^{2} – 4x + 3) (x^{2} – 4x – 12) + 54 = 0

Put x^{2} – 4x = y

(y + 3)(y – 12) + 54 = 0

y^{2} – 9y – 36 + 54 = 0

y^{2} – 9y + 18 = 0

(y – 3)(y – 6) = 0

Question 2.

Solve the equation (x – 4) (x – 2) (x – 1) (x + 1) + 8 = 0

Solution:

The equation can be rewritten as {(x – 4) (x + 1)} {(x – 2) (x – 1)} + 8 = 0

(x^{2} – 3x – 4)(x^{2} – 3x + 2) + 8 = 0