Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 3 Theory of Equations Ex 3.5 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.5

Question 1.

Solve the following equations

(i) sin² x – 5 sin x + 4 = 0

Solution:

sin^{2}x – 5sinx + 4 = 0

Let y = sin x

(y^{2} – 5y + 4 = 0

(y – 1) (y – 4) = 0

(y – 1) = o or (y – 4) = 0

y = 1 or y = 4

sin x = 1 or sin x = 4 [not possible since sin x ≤ 1]

sin x = sin \(\frac{\pi}{2}\)

x = nπ + (-1)^{n} α, n ∈ z.

x = nπ + (-1)^{n} \(\frac{\pi}{2}\)

(ii) 12x³ + 8x = 29x² – 4 = 0

Solution:

12x³ – 29x² + 8x + 4 = 0

12x² – 5x – 2 = 0

12x² – 8x + 3x – 2 = 0

4x(3x – 2) + 1(3x – 2) = 0

(3x – 2)(4x + 1) = 0

3x = 2, 4x = -1

x = \(\frac{2}{3}\) or x = –\(\frac{1}{4}\)

The roots are 2, \(\frac{2}{3}\), –\(\frac{1}{4}\).

Question 2.

Examine for the rational roots of

(i) 2x³ – x² – 1 = 0

Solution:

Sum of co-efficients = 2 – 1 – 1 = 0

⇒ x = 1 is a factor.

Which is imaginary root

∴ x = 1 is rational root.

(ii) x^{8} – 3x + 1 = 0

Solution:

a_{n} = 1; a_{0} = 1

If \(\frac{p}{q}\) is a root of the polynomial. (p, q) = 1

By rational root theorem, it has no rational roots.

Question 3.

Solve: 8x^{\(\frac{3}{2n}\)} – 8x^{\(\frac{-3}{2n}\)} = 63.

Solution:

Put k = \(\frac{3}{2n}\) ∴ 8x^{k} – 8x^{-k} = 63

8x^{k} – \(\frac{8}{x^k}\) = 63

8x^{2k} – 8 = 63x^{k}

8x^{2k} – 63x^{k} – 8 = 0

(x^{k }– 8) (8x^{k }+ 1) = 0

x^{k }– 8 = 0

x^{k }= 8

x^{\(\frac{3}{2n}\)} = 8

x = 8^{\(\frac{3}{2n}\) }(2³)^{\(\frac{3}{2n}\)} = (2²)^{n} = 4^{n}

∴ x = 4^{n} is a root of the equation.

Question 4.

Solve: 2\(\sqrt{\frac{x}{a}}\) + 3\(\sqrt{\frac{a}{x}}\) = \(\frac{b}{a}\) = \(\frac{6a}{b}\)

Solution:

put \(\sqrt{\frac{x}{a}}\) = y

Question 5.

Solve the equations

(i) 6x^{4} – 35x³ + 62x² – 35x + 6 = 0

Solution:

This is Type I even degree reciprocal equation. Hence it can be rewritten as

(1) ⇒ 6(y² – 2) – 35y + 62 = 0

6y² – 12 – 35y + 62 = 0

6y² – 35y + 50 = 0

2x² + 2 – 5x = 0 (or) 3x² + 3 = 10x

2x² – 5x + 2 = 0 (or) 3x² – 10x + 3 = 0

x = 2, \(\frac{1}{2}\) (or) x = 3, \(\frac{1}{3}\)

Roots are 2, \(\frac{1}{2}\), 3 and \(\frac{1}{3}\)

(ii) x^{4} + 3x³ – 3x – 1 = 0

Solution:

(x – 1) and (x + 1) is a factor.

(x – 1)(x + 1)(x² + 3x + 1) = 0

x – 1 = 0 (or) x + 1 = 0 (or) x² + 3x + 1 = 0

x = 1 (or) x = -1 (or) x² + 3x = -1

Question 6.

Find all real numbers satisfying

Solution:

4^{x} – 3 (2^{x+2}) + 2^{5} = 0

⇒ (2^{2})^{x} – 3(2^{x} . 2^{2}) + 2^{5} = 0

(2^{2})^{x} – 12 . 2^{x}+ 32 = 0

Let y = 2^{x}

y^{2} – 12y + 32 = 0

⇒ (y – 4) (y – 8) = 0

y – 4 = 0 or y – 8 = 0

Case (i): 2^{x} = 4

⇒ 2^{x} = (2)^{2}

⇒ x = 2

Case (ii): 2x = 8

⇒ 2^{x} = (2)^{3}

⇒ x = 3

∴ The roots are 2, 3

Question 7.

Solve the equation 6x^{4} – 5x³ – 38x² – 5x + 6 = 0 if it is known that \(\frac{1}{3}\) is a solution.

Solution:

6x^{4} – 5x³ – 38x² – 5x + 6 = 0

6(y² – 2) – 5y – 38 = 0

6y² – 12 – 5y – 38 = 0

6y² – 5y – 50 = 0

6y² – 20y + 15y – 50 = 0

2y(3y – 10) + 5(3y – 10) = 0

(3y – 10)(2y + 5) = 0

3y = 10, 2y = -5

3x² + 3 = 10x, 2x² + 2 = -5x

3x² – 10x + 3 = 0, 2x² + 5x + 2 = 0

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

x = 3, x = \(\frac{1}{3}\) or x = \(\frac{-1}{2}\), x= -2

The roots are 3, \(\frac{1}{3}\), -2, \(\frac{-1}{2}\).