You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.

Solve the following equations:

(i) sin^{2}x – 5sinx + 4 = 0

(ii) 12x^{3} + 8x = 29x^{2} – 4

Solution:

(i) sin² x – 5 sin x + 4 = 0

Put sin x = t

t² – 5t + 4 = 0

(t – 1) (t – 4) = 0

t = 1 or t = 4

sin x = 4 or sin x = 1

(is not possible) sin x = sin \(\frac{π}{2}\)

x = nπ + (-1l)^{n} \(\frac{π}{2}\) ∀ n ∈ z.

(ii) 12x^{3} + 8x = 29x^{2} – 4

12x^{3} – 29x^{2} + 8x + 4 = 0 ….. (1)

By Trail and error method, (x – 2) is a factor of (1)

The other factor is 12x^{2} – 5x – 2

The roots is 12x^{2} – 5x – 2 = 0

(3x – 2) (4x + 1) = 0

x = \(\frac{2}{3}\), x = \(-\frac{1}{4}\)

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

[Here a_{n} = 12, a_{0} = 4; Let \(\frac{p}{q}\) be the root of the equation (1)

The factors of a_{0 }: ±1, ±2, ±4 (P must divisible by 4)

The factor of a_{n} : ±1, ±2, ±3, ±4, ±6, ±12

q must divide as (12)

Using these p and q we can form \(\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{2}{3}, \pm \frac{1}{6}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm 3\) are the possible roots of equation. (1)]

Question 2.

Examine for the rational roots of:

(i) 2x^{3} – x^{2} – 1 = 0

(ii) x^{8} – 3x + 1 = 0

Solution:

(i) 2x^{3} – x^{2} – 1 = 0

Sum of the co-efficients = 0

∴ (x – 1) is a factor

The other factor is 2x^{2} + x + 1.

The root is (2x^{2} + x + 1) = 0

Here ∆ = b^{2} – 4ac = (1)^{2} – 4(2) (1) = 1 – 8 = -7 < 0

The remaining roots are imaginary.

The only rational root is x = 1

(ii) x^{8} – 3x + 1 = 0 …. (1)

Here a_{n} = 1, a_{0} = 1

If \(\frac{p}{q}\) is a rational root of (1)

Then q is a factor a_{n}, p is a factor of a_{0}

The possible values of p and q are ± 1.

Among the possible values 1, -1, [(p, q) = 1]

None of them satisfies the equation (1)

The above equation has no rational roots.

Question 3.

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

Solution:

Squaring on both sides

only possible solution is x = 4^{n}

Question 4.

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

Solution:

Question 5.

Solve the equations:

(i) 6x^{4} – 35x^{3} + 62x^{2} – 35x + 6 = 0

(ii) x^{4} + 3x^{3} – 3x – 1 = 0

Solution:

(i) 6x^{4} – 35x^{3} + 62x^{2} – 35x + 6 = 0 ….. (1)

It is a even degree reciprocal equation as p(x) = \(x^{n} p\left(\frac{1}{x}\right)\)

Dividing equation (1) by x^{2},

(ii) x^{4} + 3x^{3} – 3x – 1 = 0 …… (1)

It is an even degree reciprocal function of type II.

1, -1 are the solution of equation (1)

(x – 1), (x + 1) are the factor of (1)

(x^{2} – 1) is a factor of (1)

Dividing (1) by (x^{2} – 1)

we get, x^{2} + 3x + 1 = 0 is the other factor.

Question 6.

Find all real numbers satisfying 4^{x} – 3(2^{x+2}) + 2^{5} = 0

Solution:

4^{x} – 3(2^{x+2}) + 2^{5} = 0.

(2^{x})² – 3.(2^{x }2²) + 2^{5} = 0

(2^{x})² – 12(2^{x}) + 32 = 0

Put 2^{x }= t

(t² – 12t + 32 = 0)

(y – 4)(y – 8) = 0

y = 4 or y = 8

t = 8 (or) t = 4

2^{x} = 8 = 2³ (or) 2^{x} = 4 = (2)²

x = 3 (or) x = 2

Roots are 3, 2

Question 7.

Solve the equation 6x^{4} – 5x^{3} – 38x^{2} – 5x + 6 = 0 if it is known that \(\frac{1}{3}\) is a solution.

Solution:

6x^{4} – 5x^{3} – 38x^{2} – 5x + 6 = 0 …… (1)

x = \(\frac{1}{3}\) is a Solution

∴ (3x – 1) is a factor of (1)

(1) is a Reciprocal equation even degree divide (1) by x^{2}.

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

Question 1.

Solve: 3^{2x + 4} + 1 = 2.3^{x + 2}

Solution:

3^{2x + 4} = 3^{x + 2} + 3^{x + 2} – 1

3^{2x + 4} – 3^{x + 2} = 3^{x + 2} – 1 ⇒ 3^{x + 2}[3^{x + 2} – 1] = [3^{x + 2} – 1]

3^{x + 2}= 1 ⇒ 3^{x + 2} = 3^{0}

x + 2 = 0 ⇒ x = -2

Question 2.

Solve: 2^{x} – 2^{x + 3} + 2^{4} = 0.

Solution:

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

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

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

2^{x} = 2^{2} ⇒ x = 2

Question 3.

Solve: (x – 4) (x + 2) (x + 3) (x – 3) + 8 = 0.

Solution:

(x – 4) (x + 3) (x + 2) (x – 3) + 8 = 0

(x^{2} – x – 12) (x^{2} – x – 6) + 8 = 0

Let y = x^{2} – x

(y – 12) (y – 6) + 8 = 0 ⇒ y^{2} – 18y + 72 + 8 = 0

y^{2} – 18y + 80 = 0 ⇒ (y – 10)( y – 8) = 0

Case (i) y – 10 = 0

x^{2} – x – 8 = 0

Question 4.

Solution:

Question 5.

5^{x – 1} + 5^{1 – x} = 26

Solution:

Question 6.

Solve: 12x^{4} – 56x^{3} + 89x^{2} – 56x + 12 = 0

Solution:

Since the coefficients of the equations are equal from both ends.

Divide the equation by x^{2}

Question 7.

Solution:

Question 8.

Solve: x^{4} + 4x^{3} + 5x^{2} + 4x + 1 = 0

Solution: