# Exercise 6

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

(a) if a is greater than b

(b) if a is smaller than b

(c) if a is equal to b

(d) if a is either equal to or greater than b

(e) if a is either equal to or smaller than b

Question.1
I. $\sqrt{2304}$ = a
II. b2 = 2304

#### View Ans & Explanation

Ans.c

From I : If $\sqrt{2304}$ = a

then a = ±48

(Do not consider – 48 as value of a)

Again,

From II :

If b2 = 2304 then b = +48

Hence a = b.

Question.2
I. 12a2 - 7a + 1 = 0
II. 15b2 - 16b + 4 = 0

#### View Ans & Explanation

Ans.b

I. 12a2 – 7a + 1 = 0

II. 15b2 – 16b + 4 = 0

Sum of the two values of a,i.e., (a1, + a2)

= -(-7)12 = 712

Similarly,

Sum of the two values of b,

i.e., (b1 + b2) = -(-16)15 = 1615

since (7)12 < 1615

Therefore, a < b,

Now check the equality of root

(12 × 4 - 15 × 1)2 = {12 × (-16) - 15 × (-7)}

{(-7) × 4 - (-16) × 1}

⇒ 332 = {-87}{-12}

⇒ 1089 = 1044, which is not true.

Therefore, out answer should be a < b.

Question.3
I. a2 + 9a + 20 = 0
II. 2b2 - 10b + 12 = 0

#### View Ans & Explanation

Ans.b

I. a2 + 9a + 20 = 0

Break 9 as F1 and F2, so that F1 × F2 = 20 and F1 + F2 = 9.

Therefore, F1 = 5, F2 = 4

Now one value of a = -5/1 = -5

other value of a = -20/5 = -4

II. 2b2 + 10b + 12 = 0

The two parts of 10, ie F1 = 6 and F2 = 4

∴ Value of b = -62 = -3 and -126 = -2

Obviously b > a.

If general form of quadratic equation is

ax2 + bx + c = 0,

then split b into two parts so that b1 + b2 = b and b1 × b2e = a × c

Now say b1 as F1 and b2 as F2. Then the values of 'x' will be -F1a and -CF1 or -F2a and -CF2

Question.4
I. 3a + 2b = 14
II. a + 4b – 13 = 0

#### View Ans & Explanation

Ans.a

I. 3a + 2b = 14

II. a + 4b = 13

Substract equation I from equation II after multiplying II by 3.

We get 3a + 12b – 3a – 2b = 39 – 14

⇒ 10b = 25

⇒ b = 2.5

Put value of b in equation II. We set a + 4 × 2.5 = 13.

Therefore, a = 3. Thus, a > b

Question.5
I. a2 – 7a + 12 = 0
II. b2 – 9b + 20 = 0

#### View Ans & Explanation

Ans.e

I. a2 – 7a + 12 = 0

Here, F1 = – 4 and F2 = –3

Now, values of a = -(-4)1 = 4

and -12-4 = 3

II. b2 – 9b + 20 = 0

Here F1 = – 5 and F2 = – 4

Now, values of a = -(-5)1 = 5

And -20-5 = 4

Thus b ≥ a

Directions (Qs. 6 - 10): In each question one or more equation(s) is(are) provided. On the basis of these you have

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.6
528 × 98p = 1514 × 1316q

#### View Ans & Explanation

Ans.b

528 × 98p = 1514 × 1316q

or, 45p224 = 195q224

or, 3p = 13q

∴ p > q

Question.7
(i) p – 7 = 0
(ii) 3q2 – 10q + 7 = 0

#### View Ans & Explanation

Ans.b

(i) p – 7 = 0 or, p = 7

(ii) 3q2 – 10q + 7 = 0

or, 3q2 – 3q – 7q + 7 = 0

or, 3q(q – 1) – 7(q – l) = 0

or, (3q – 7)(q – 1) or, q = 1 or,73

\ p > q

Question.8
(i) 4p2 = 16
(ii) q2 – 10q + 25 = 0

#### View Ans & Explanation

Ans.c

(i) 4p22 = 16; p = √4 = 2

(ii) q2 – l0q + 25 = 0

⇒ (q – 5)(q – 5) = 0

or, q = 5 \ q > p

Question.9
(i) 4p2 – 5p + 1 = 0
(ii) q2 – 2q + 1 = 0

#### View Ans & Explanation

Ans.e

(i) 4p2 – 5p + 1 = 0

or, 4p2 – 4p – p + i = 0

or, 4p(p – 1) – 1(p – 1) = 0

or, (4p – 1)(p – 1) = 0

or, p = 1 and p = 14

(ii) q2 – 2q + 1 = 0

(q – 1)(q – 1) = 0

or, q = 1

\ q ³ p

Question.10
(i) q2 – 11q + 30 = 0
(ii) 2p2 – 7p + 6 = 0

#### View Ans & Explanation

Ans.c

q = 5, 6 & p = 32 , 2