Students must practice this TS Intermediate Maths 2A Solutions Chapter 3 Quadratic Expressions Ex 3(c) to find a better approach to solving the problems.

## TS Inter 2nd Year Maths 2A Solutions Chapter 3 Quadratic Expressions Exercise 3(c)

I.

Question 1.

Solve the following inequatlons by algebraic method.

i) 15x^{2} – 4x – 4 ≤ 0

ii) x^{2} – 2x + 1 < 0

iii) 2 – 3x – 2x^{2} ≥ 0

iv) x^{2} – 4x – 21 ≥ 0

Solution:

i) 15x^{2} – 4x – 4 ≤ 0

15x^{2} + 10x – 6x – 4 ≤ 0

5x (3x + 2) – 2 (3x + 2) ≤ 0

(3x + 2) (5x – 2) ≤ 0

\(\frac{-2}{3}\) ≤ x ≤ \(\frac{2}{5}\).

ii) x^{2} – 2x + 1 < 0

(x- 1)^{2} < 0

Not possible

∵ (x – 1)^{2} ≥ 0

No solution.

iii) 2 – 3x – 2x^{2} ≥ 0

2x^{2} + 3x – 2 ≤ 0

2x^{2} + 4x – x – 2 ≤ 0

2x (x + 2) – 1 (x + 2) ≤ 0

(2x – 1)(x + 2) ≤ 0

– 2 ≤ x ≤ \(\frac{1}{2}\)

iv) x^{2} – 4x – 21 ≥ 0

x^{2} – 7x + 3x – 21 ≥ 0

x (x – 7) + 3 (x – 7) ≥ 0

(x + 3) (x – 7) ≥ 0

x ≥ 7 or x ≤ – 3

(- ∞ < x ≤ – 3) ∪ (7 ≤ x < ∞).

II.

Question 1.

Solve the following inequations by graphical method.

i) x^{2} – 7x + 6 > 0

ii) 4 – x^{2} > o

iii) 15x^{2} + 4x – 4 ≤ 0

iv) x^{2} – 4x – 21 ≥ 0

Solution:

i) (x – 6) (x – 1) > 0

ii) 4 – x^{2} > 0

x^{2} – 4 > 0

(x – 2) (x + 2) > 0

– 2 < x < 2

iii) 15x^{2} + 4x – 4 ≤ 0

15x^{2} + 10x – 6x – 4 ≤ 0

5x (3x + 2) – 2 (3x + 2) ≤ 0

\(\frac{-2}{3} \leq x \leq \frac{2}{5}\)

iv) x^{2} – 4x – 21 ≥ 0

(x – 7) (x + 3) ≥ 0

x ≥ 7 or x ≤ – 3

Question 2.

Solve the following Inequations.

i) \(\sqrt{3 x-8}\) < – 2

ii) \(\sqrt{-x^2+6 x-5}\) > 8 – 2x

Solution:

\(\sqrt{3 x-8}\) < – 2 Possible when 3x – 8 > 0

x > \(\frac{8}{3}\)

also \(\sqrt{3 x-8}\) ≥ 0

∴ Solution does not exist.

ii) \(\sqrt{-x^2+6 x-5}\) > 8 – 2x

Possible

– x^{2} + 6x – 5 ≥ 0

x^{2} – 6x + 5 ≤ 0

(x – 5) (x – 1) ≤ 0

1 ≤ x ≤ 5 …………….(1)

Squaring on both sides we get

– x^{2} + 6x – 5 > 64 + 4x^{2} – 32x

0 > 5x^{2} – 38x + 64 + 5

or 5x^{2} – 38x + 69 < 0

5x^{2} – 23x – 15x + 69 < 0

5x (x – 3) – 23(x – 3)< 0

(x – 3) (5x – 23) < 0