# Exercise 2

Directions (Qs.1-5): For the two given equations I and II give answer

(a) if p is greater than q

(b) if p is smaller than q

(c) if p is equal to q.

(d) if p is either equal to or greater than q

(e) if p is either equal to or smaller than q.

Question.1
I. 6p2 + 5p + 1 = 0
II. 20q2 + 9q = –1

#### View Ans & Explanation

Ans.b

I. 6p2 + 5p + 1 = 0

or,6p2 + 3p + 2p + 1 = 0

or,3p(2p+ 1) + 1 (2p + 1) = 0

or,(3p + 1)(2p + 1) = 0

Hence, p = -13 , -12

II. 20q2 + 9q + 1 = 0

or, 20q2 + 5q + 4q + 1 = 0

or, 5q(4q + l) + 1(4q + 1) = 0

or,(5q + 1)(4q + l) =0

Hence, q = -15 , -14

Thus, p < q.

Question.2
I. 3p2 + 2p – 1 = 0
II. 2q2 + 7q + 6 = 0

#### View Ans & Explanation

Ans.a

I. 3p2 + 2p – 1 = 0

or, 3p2 + 3p – p – 1 = 0

or, 3p(p + l) – 1(p + 1) = 0

or,(3p – l)(p + 1) = 0

Therefore, p = 13 , -1

II. 2q2 + 7q + 6 = 0

or,2q2 + 4q + 3q + 6 = 0

or, 2q(q + 2) + 3(q + 2) = 0

or,(2q + 3)(q + 2) = 0

Therefore, q = -32, -2

thus p > q

Question.3
I. 3p2 + 15p = –18
II. q2 + 7q + 12 = 0

#### View Ans & Explanation

Ans.d

I. 3p2 + 15p + 18 =0

or,3p2 + 9p + 6p + 18 = 0

or,3p(p+ 3) + 6(p + 3) = 0

or,(3p + 6)(p + 3) = 0

or p = -63,3 = -2 , 3

II. q2 + 7q + 12 = 0

or, q2 + 4q + 3q + 12 = 0

or, q(q + 4) + 3(q + 4) = 0

or, (q + 3)(q + 4) = 0

or, q = – 3, –4

Therefore, p ≥ q

Question.4
I. p = √4√9
II. 9q2 – 12q + 4 = 0

#### View Ans & Explanation

Ans.c

I. p = √4√9 = 23

II. 9q2 – 12q + 4 = 0

or, 9q2 – 6q – 6q + 4 = 0

or, 3q(3q – 2) – 2(3q – 2) = 0

or, (3q – 2)(3q – 2) = 0

or, q = 23

Therefore, p = q

Question.5
I. p2 + 13p + 42 = 0
II. q2 = 36

#### View Ans & Explanation

Ans.e

I. p2 + 13p + 42 = 0

or, p2 + 7p + 6p + 42 = 0

or, p(p + 7) + 6(p + 7) = 0

or, (p + 6)(p + 7) = 0

or, p = – 6, –7

II. q2 = 36

q = √36

q = +6, –6

Therefore, p ≤ q.