Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(c) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2A Quadratic Expressions Solutions Exercise 3(c)

I.

Question 1.

Solve the following inequations by the algebraic method.

(i) 15x^{2} + 4x – 4 ≤ 0

Solution:

15x^{2} + 4x – 4 ≤ 0

⇒ 15x^{2} – 6x + 10x – 4 ≤ 0

⇒ 3x(5x – 2) + 2(5x – 2) ≤ 0

⇒ (3x + 2) (5x – 2) ≤ 0

Co-efficient of x^{2} = 15 > 0,

Given Expression is ≤ 0

⇒ x lies between \(\frac{-2}{3}\) and \(\frac{2}{5}\)

i.e., \(\frac{-2}{3} \leq x \leq \frac{2}{5}\)

(ii) x^{2} – 2x + 1 < 0

Solution:

x^{2} – 2x + 1 < 0

⇒ (x – 1)^{2} < 0

There is no real value of ‘x’ satisfying this inequality

Solution set = Φ (or) Solution does not exist.

(iii) 2 – 3x – 2x^{2} ≥ 0

Solution:

-(2x^{2} + 3x – 2) ≥ 0

⇒ -(2x^{2} + 4x – x – 2) ≥ 0

⇒ -[2x(x + 2) – 1(x + 2)] ≥ 0

⇒ -(2x – 1) (x + 2) ≥ 0

Co-efficient of x^{2} = -2 < 0,

The given expression is ≥ 0

⇒ x lies between -2 and \(\frac{1}{2}\)

i.e., -2 ≤ x ≤ \(\frac{1}{2}\)

(iv) x^{2} – 4x – 21 ≥ 0

Solution:

x^{2} – 4x – 21 ≥ 0

⇒ x^{2} – 7x + 3x – 21 ≥ 0

⇒ x(x – 7) + 3(x – 7) ≥ 0

⇒ (x + 3) (x – 7) ≥ 0

Co-efficient of x^{2} = 1 > 0,

The given expression is ≥ 0

x does not lie between -3 and 7

i.e., {x/x ∈ (-∞, -3] ∪ [7, ∞)}

II.

Question 1.

Solve the following inequations by graphical method.

(i) x^{2} – 7x + 6 > 0

Solution:

f(x) = x^{2} – 7x + 6

f(x) > 0 ⇒ y > 0

Solutions are given by x < 1 and x > 6

(ii) 4 – x^{2} > 0

Solution:

Let f(x) = 4 – x^{2}

f(x) > 0 ⇒ y > 0

Solution set = {x/-2 < x < 2}

(iii) 15x^{2} + 4x – 4 < 0

Solution:

Let f(x) = 15x^{2} + 4x – 4

f(x) ≤ 0 ⇒ y ≤ 0

Solution set = \(\left\{x / \frac{-2}{3} \leq x \leq \frac{2}{5}\right\}\)

(iv) x^{2} – 4x – 21 ≥ 0

Solution:

Let f(x) = x^{2} – 4x – 21

f(x) ≥ 0 ⇒ y ≥ 0

Solution set = {x/x ∈ (-∞, -3] ∪ [7, ∞)}

Question 2.

Solve the following inequations.

(i) \(\sqrt{3 x-8}\) < -2

Solution:

L.H.S. is positive and R.H.S. is negative.

∴ The given inequality holds for no real x.

Solution set = Φ (or) Solution does not exist.

(ii) \(\sqrt{-x^{2}+6 x-5}\) > 8 – 2x

Solution:

\(\sqrt{-x^{2}+6 x-5}\) > 8 – 2x

⇔ -x^{2} + 6x – 5 > 0

and (i) 8 – 2x < 0 (or) (ii) 8 – 2x ≥ 0

We have -x^{2} + 6x – 5 = -(x^{2} – 6x + 5) = -(x – 1) (x – 5)

Hence -x^{2} + 6x – 5 ≥ 0 ⇔ x ∈ [1, 5]

(i) -x^{2} + 6x – 5 ≥ 0 and 8 – 2x < 0

⇔ x ∈ [1, 5] and x > 4

⇔ x ∈ [4, 5] ………(1)

(ii) -x^{2} + 6x – 5 ≥ 0 and 8 – 2x ≥ 0

∵ \(\sqrt{\left(-x^{2}+6 x-5\right)}\) > 8 – 2x

⇔ -x^{2} + 6x – 5 > (8 – 2x)^{2} and 8 – 2x ≥ 0

⇔ -x^{2} + 6x – 5 > 64 + 4x^{2} – 32x and x ≤ 4

⇔ -5x^{2} + 38x – 69 > 0 and x ≤ 4

⇔ 5x^{2} – 38x + 69 < 0 and x ≤ 4

⇔ 5x^{2} – 15x – 23x + 69 < 0 and x ≤ 4

⇔ (5x – 23)(x – 3) < 0 and x ≤ 4

⇔ x ∈ (3, \(\frac{23}{5}\)) and x ≤ 4

⇔ x ∈ (3, \(\frac{23}{5}\)) ∩ (-∞, 4)

⇔ x ∈ (3, 4)

Hence the solution set of the given equation is x ∈ (4, 5) ∪ (3, 4)

⇒ x ∈ (3, 5) (or) 3 < x ≤ 5.