Exercise 4


Directions (Qs.1-5): In each of these questions two equations are given. You have to solve these equations and 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 – 6x = 7
II. 2y2 + 13y + 15 = 0
Ans.b

I. x2 – 6x = 7

or, x2 – 6x – 7 = 0

or, (x – 7)(x + 1) = 0

or, x = 7, –1

II. 2y2 + 13y + 15 = 0

or, 2y2 + 3y + 10y + 15 = 0

or, (2y + 3)(y + 5) = 0 or, y = -32 , -5

Hence, x > y

Question.2
I. 3x2 – 7x + 20
II. 2y2 – 11y + 15 = 0
Ans.a

I. 3x2 – 7x + 2 = 0

or, 3x2 – 6x – x + 2 = 0

or, (x – 2)(3x – 1) = 0

or, x = 2, 1/3

II. 2y2 – 11y + 15 = 0

or, 2y2 – 6y – 5y + 15 = 0

or, (2y – 5)(y – 3) = 0

or, y = 5/2, 3

Hence, y > x

Question.3
I. 10x2 – 7x + 1 = 0
II. 35y2 – 12y + 1 = 0
Ans.d

I. 10x2 – 7x + 1 = 0

or, 10x2 – 5x – 2x + 1 = 0

or, (2x – 1)(5x – 1) = 0

or, x = 12, 15

II. 35y2 – 12y + 1 = 0

or, 35y2 – 7y – 5y + 1 = 0

or, (5y – 1)(7y – 1) = 0

or, y = 15 , 17

Hence, x ≥ y

Question.4
I. 4x2 = 25
II. 2y2 – 13y + 21 = 0
Ans.a

I. 4x2 = 25

or x = ±52

II. 2y2 – 13y + 21 = 0

or, 2y2 – 6y – 7y + 21 = 0

or, (y – 3)(2y – 7) = 0

or, y = 3, 72

Hence, y > x

Question.5
I. 3x2 + 7x = 6
II. 6(2y2 + 1) = 17y
Ans.e

I. 3x2 + 7x – 6 = 0

or, 3x2 + 9x – 2x – 6 = 0

or, (x + 3)(3x – 2) = 0

or, x = – 3, 23

II. 6(2y2 + 1) = 17y

or, 12y2 + 6 – 17y = 0

or, 12y2 – 9y – 8y + 6 = 0

or, (4y – 3)(3y – 2) = 0

or, y = 34 , 23

Hence, y ≥ x

Directions (Qs. 6-8): In each question below one or more equation(s) is/are given. On the basis of these, you have to find out the relationship between p and q.

Give answer (a) if p = q

Give answer (b) if p > q

Give answer (c) if p < q

Give answer (d) if p ≤ q

Give answer (e) if p ³ q

Question.6
I. 2p2 = 23p – 63
II. 2q(q-8) = q-36
Ans.b

I. 2p2 = 23p – 63

or, 2p2 – 23p + 63 = 0

II. 2q(q-8) = q-36

or, (2p – 9)(p – 7) = 0

or, q-7 × q36 = 12

\p = 92, or 7

or, q29 = 12

\ q = \(\left [ \frac{1}{2} \right ]^{\frac{1}{29}}\)

Hence, p > q

Question.7
I. p(p-1) = (p-1)
II. q2 = 4q-1
Ans.c

I. p(p-1) = p-1

II. q2 = 4(q-1)

or, p × 1p = 1p

\p = 1

or, q3 = 4

\q = (4)13 > 1

Hence, q > p

Question.8
I. 2p(p – 4) = 8(p + 5)
II. q2 + 12 + 7q
Ans.b

I. 2p(p – 4) = 8(p + 5)

or p2 – 8p – 20 = 0

or (p + 2)(p – 10) = 0

⇒ p = –2, or + 10

II. q2 + 7q + 12 = 0

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

q = –3 or –4

p > q