# Exercise 7

Directions (Qs. 1 - 3) : In each question below one or more equation(s) is/are provided. On the basis of these, you have to find out relation between p and q.

Give answer (a) if p = q,

Give answer (b) if p > q,

Give answer (c) if q > p,

Give answer (d) if p ≥ q and

Give answer (e) if q ≥ p.

Question.1
I. 2p2 + 40 = 18p
II. q2 = 13q – 42

#### View Ans & Explanation

Ans.c

I. 2p2 + 40 = 18p

or, p2 – 9p + 20 = 0

or, (p – 4)(p – 5) = 0

∴ p = 4 or 5

II. q2 = 13q – 42

or, q2 – 13q + 42 = 0

or, (q – 7) (q – 6) = 0

∴ q = 6 or 7

Hence, q > p

Question.2
I. 4q2 + 8q = 4q + 8
II. p2 + 9p = 2p – 12

#### View Ans & Explanation

Ans.c

I. 4q2 + 8q = 4q + 8

or, q2 + q – 2 = 0

or, (q – 1)(q + 2) = 0

∴ q = 1 or – 2

II. p2 + 9p = 2p – 12

or, p2 + 7p + 12 = 0

or, (p + p)(p + (c) = 0

∴ p = –3 or –4

Hence, q > p

Question.3
I. 6q2 + 12 = 72q
II. 12p2 + 2 = 10p

#### View Ans & Explanation

Ans.d

I. 6q2 + 12 = 72q

or, 12q2 – 7q + 1 = 0

or (q - 14)(q - 13) = 0

∴ q = 14 or 13

II. 12p2 + 2 = 10p

or, 6p2 – 5p + 1 = 0

or (p - 13)(p - 12) = 0

∴ p = 12 or 13

Hence, p ≥ q

Directions (Qs. 4 - 8): In each of the following questions two equations are given. You have to solve them and give answer

(a) if x > y;

(b) if x < y;

(c) if x = y;

(d) if x ≥ y;

(e) if x ≤ y;

Question.4
I. y2 – 6y + 9 = 0
II. x2 + 2x – 3 = 0

#### View Ans & Explanation

Ans.b

I. y2 – 6y + 9 = 0

or, (y – 3)2 = 0 or, y = 3

II. x2 + 2x – 3 = 0

or, x = 1, – 3

Hence, y > x

Question.5
I. x2 – 5x + 6 = 0
II. 2y2 + 3y – 5 = 0

#### View Ans & Explanation

Ans.a

I. x2 – 5x + 6 = 0

or, (x – 3)(x – 2) = 0

or, x = 2, 3

II. 2y2 – 3y – 5 = 0

or, y = 1, 52

∴ Hence, x > y

Question.6
I. x = $\sqrt{256}$

II. y = (– 4)2

#### View Ans & Explanation

Ans.c

I. x = $\sqrt{256}$ = 16

II. y = (– 4)2 = 16

Hence, x = y

Question.7
I. x2 – 6x + 5 = 0
II. y2 – 13y + 42 = 0

#### View Ans & Explanation

Ans.b

I. x2 – 6x + 5 = 0

or, x = 1, 5

II. y2 – 13y + 42 = 0

or, (y – 7)(y – 6) = 0

or, y = 6, 7

Hence, y > x

Question.8
I. x2 + 3x + 2 = 0
II. y2 – 4y + 1 = 0

#### View Ans & Explanation

Ans.b

I. x2 + 3x + 2 = 0

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

or, x = – 2 or, –1

II. y2 – 4y + 1 = 0 or, y = 2 ± √3

Hence, y > x