Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.4 Textbook Questions and Answers, Notes.

## 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

Solution:

(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 – 2)(x- 7)(x + 1) = 16

Solution:

(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

Question 2.

Solve:

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

Solution:

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

(x² + x – 6) (4x² + 6x – 2x – 3) + 20 = 0

(x² + x – 6) (4x² + 4x – 3) + 20 = 0

(x² + x – 6) (x² + 4x – \(\frac{3}{4}\)) + 20 = 0

(÷4) (x² + x – 6) (x² + x – \(\frac{3}{4}\)) + \(\frac{20}{4}\) = 0

Put y = x² + x

4y² – 27y – 38 = 0

4y² – 8y – 19y + 38 = 0

(4y – 19) (y – 2) = 0

y = 2 (or) y = \(\frac{19}{4}\)

x² + x = 2 (or) x² + x = \(\frac{19}{4}\)

x² + x – 2 = 0 (or) 4x² + 4x = 19

(x + 2)(x – 1) = 0 (or) 4x² + 4x – 19 = 0

x = -2 or x = 1