# Exercise 3

Directions (Qs. 1 - 4): In each of the following questions, one or two equation(s) is/are given. On their basis you have to determine the relation between x and y and then give answer

(a) if x < y

(b) if x > y

(c) if x ≤ y

(d) if x ≥ y

(e) if x = y

Question.1
I. x2 + 3x + 2 = 0
II. 2y2 = 5y

#### View Ans & Explanation

Ans.a

I. x2 + 3x + 2 = 0

or, x2 + 2x + x + 2 = 0

or,(x + 2)(x + 1) = 0

or, x = –2, – 1

II. 2y2 = 5y

or, 2y2 – 5y = 0

or, y(2y – 5) = 0

or, y = 0 , 52

Hence, y> x

Question.2
I. 2x2 + 5x + 2 = 0
II. 4y2 = 1

#### View Ans & Explanation

Ans.c

I. 2x2 + 5x + 2 = 0

or, 2x2 + 4x + x + 2 = 0

or,(x + 2)(2x + 1) = 0

or, x = –2, -12

II. 4y2 = 1

or y2 = 14

or, y = ± 12

Hence, y ≥ x

Question.3
I. y2 + 2y – 3 = 0
II. 2x2 – 7x + 6 = 0

#### View Ans & Explanation

Ans.b

I. y2 + 2y – 3 = 0 ⇒ (y – 1)(y + 3) = 0

or, y = 1, –3

II. 2x2 – 7x + 6 = 0

or, 2x2 – 4x – 3x + 6 = 0

or, (2x – 3)(x – 2) = 0

or, x = 2, 3/2

Hence, x > y

Question.4
I. x2 – 5x + 6 = 0
II. y2 + y – 6 = 0

#### View Ans & Explanation

Ans.d

I. x2 – 5x + 6 = 0

or, x2 – 3x – 2x + 6 = 0

or, x(x – 3) – 2(x – 3) = 0

or,(x – 3)(x – 2) = 0

or, x = 2, 3

II. y2 + y – 6 = 0

or, y2 + 3y – 2y – 6 = 0

or, y(y + 3) – 2 (y + 3) = 0

or, (y + 3)(y – 2) = 0

or, y = 2, –3

Hence, x > y

Directions (Qs. 5 - 8) : In each of the following questions two equations are given. You have to solve them and Give answer

(a) if p < q (b) if p > q

(c) if p ≤ q

(d) if p ≥ q

(e) if p = q

Question.5
I. p2 – 7p = – 12
II. q2 – 3q + 2 = 0

#### View Ans & Explanation

Ans.b

I. p2 – 7p = – 12

or, p2 – 7p + 12 = 0

or, (p – 3)(p – 4) = 0

or, p = 3 or 4

II. q2 – 3q + 2 = 0

or, (q – 2)(q– 1) = 0

or, q = 1 or 2

Hence, p > q

Question.6
I. 12p2 – 7p = – 1
II. 6q2 – 7q + 2 = 0

#### View Ans & Explanation

Ans.a

I. 12p2 – 7p = – 1

or, 12p2 – 7p + 1= 0

or, (3p – 1)(4p – 1) = 0

or, p = 14 or 13

II. 6q2 – 7q + 2 = 0

or, (3q – 2)(2q – 1) = 0

or, q = 12 or 23

Hence, q > p

Question.7
I. p2 – 8p + 15 = 0
II. q2 – 5q = –6

#### View Ans & Explanation

Ans.d

I. p2 – 8p + 15 = 0

or, (p – 3)(p – 5) = 0

or, p = 3 or 5

II. q2 – 5q + 6 = 0

or, (q – 2)(q – 3) = 0

or, q = 2 or 3

Hence, p > q

Question.8
I. 2p2 + 20p + 50 = 0
II. q2 – 25

#### View Ans & Explanation

Ans.c

I. 2p2 + 20p + 50 = 0

or, p2 + 10p + 25 = 0

or, (p + 5)2 = 0

or, p = –5

II. q2 = 25

or, q = ±5

Hence, p ≤ q