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

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

Question 1.
Solve the following equations:
(i) sin²x – 5sinx + 4 = 0
(ii) 12x³ + 8x = 29x² – 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) 12x³ + 8x = 29x² – 4
12x³ – 29x² + 8x + 4 = 0 ….. (1)
By Trail and error method, (x – 2) is a factor of (1)
The other factor is 12x² – 5x – 2
The roots is 12x² – 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, a₀ = 4; Let \(\frac{p}{q}\) be the root of the equation (1)
The factors of a₀: ±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) 2x³ – x² – 1 = 0
(ii) x⁸ – 3x + 1 = 0
Solution:
(i) 2x³ – x² – 1 = 0
Sum of the co-efficients = 0
∴ (x – 1) is a factor
The other factor is 2x² + x + 1.
The root is (2x² + x + 1) = 0
Here ∆ = b² – 4ac = (1)² – 4(2) (1) = 1 – 8 = -7 < 0
The remaining roots are imaginary.
The only rational root is x = 1

(ii) x⁸ – 3x + 1 = 0 …. (1)
Here a_n = 1, a₀ = 1
If \(\frac{p}{q}\) is a rational root of (1)
Then q is a factor a_n, p is a factor of a₀
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 = 4ⁿ

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) 6x⁴ – 35x³ + 62x² – 35x + 6 = 0
(ii) x⁴ + 3x³ – 3x – 1 = 0
Solution:
(i) 6x⁴ – 35x³ + 62x² – 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 x²,

(ii) x⁴ + 3x³ – 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)
(x² – 1) is a factor of (1)
Dividing (1) by (x² – 1)
we get, x² + 3x + 1 = 0 is the other factor.

Question 6.
Find all real numbers satisfying 4^x – 3(2^x+2) + 2⁵ = 0
Solution:
4^x – 3(2^x+2) + 2⁵ = 0.
(2^x)² – 3.(2^x2²) + 2⁵ = 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 6x⁴ – 5x³ – 38x² – 5x + 6 = 0 if it is known that \(\frac{1}{3}\) is a solution.
Solution:
6x⁴ – 5x³ – 38x² – 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 x².

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} = 3⁰
x + 2 = 0 ⇒ x = -2

Question 2.
Solve: 2^x – 2^{x + 3} + 2⁴ = 0.
Solution:
2^2x – (2^x.2³) + 2⁴ = 0 since 2³ = 8 = 4 + 4 = 2² + 2²
2^2x – (4 + 4)2^x + 2⁴ = 0 ⇒ 2^2x – (2² + 2²)2^x + 2⁴ = 0
(2^x – 2²)(2^x – 2²) = 0 ⇒ (2^x – 2²)² = 0
2^x = 2² ⇒ 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
(x² – x – 12) (x² – x – 6) + 8 = 0
Let y = x² – x
(y – 12) (y – 6) + 8 = 0 ⇒ y² – 18y + 72 + 8 = 0
y² – 18y + 80 = 0 ⇒ (y – 10)( y – 8) = 0
Case (i) y – 10 = 0
x² – x – 8 = 0

Question 4.

Solution:

Question 5.
5^{x – 1} + 5^{1 – x} = 26
Solution:

Question 6.
Solve: 12x⁴ – 56x³ + 89x² – 56x + 12 = 0
Solution:
Since the coefficients of the equations are equal from both ends.
Divide the equation by x²