Directions: 1-5) In the following questions, two equations numbered I and II are given with variables a and b. Solve both the equations and find out the relationship between a and b. Then give answer accordingly.

1.
I. 3a2 – a – 10 = 0
II. 3b2 – 11b + 6 = 0

Ans:5
3a2 – a – 10 = 03a2 – 6a + 5a – 10 = 0
3a (a-2) + 5 (a-2) = 0
(3a+5) (a – 2) = 0
Gives a = -5/3, 2
3b2 – 11b + 6 = 0
3b2 – 9b – 2b +6 =0
3b (b-3) – 2(b-3) =0
(3b -2) (b-3) =0
Gives b = 2/3, 3
Relationship cannot be determined

2.
I. 3a2 – (3 – 2√2)a – 2√2 = 0
II. 3b2 – (1 + 3√3)b + √3 = 0

Ans:1
3a2 – (3 – 2√2)a – 2√2 = 0
(3a2 – 3a) + (2√2a – 2√2) = 0
3a (a – 1) + 2√2 (a – 1) = 0
So a = 1, -2√2/3 (-0.9)
3b2 – (1 + 3√3)b + √3 = 0
(3b2 – b) – (3√3b – √3) = 0
b (3b – 1) – √3 (3b – 1) = 0
So, b = 1/3, √3 (1.7)Relationship cannot be determined

3.
I. a2+ (4 + √2)a + 4√2 = 0
II. 5b2+ (2 + 5√2)b + 2√2 = 0

Ans:4
a2 + (4 + √2)a + 4√2 = 0
(a2 + 4a) + (√2a + 4√2) = 0
a (a + 4) + √2 (a + 4) = 0
So a = -4, -√2 (-1.4)
5b2 + (2 + 5√2)b + 2√2 = 0
(5b2 + 2b) + (5√2b + 2√2) = 0
b (5b + 2) + √2 (5b + 2) = 0
So, b = -2/5 (-0.4), -√2 (-1.4)a ≤ b

4.
I. 6a2– (3 + 4√3)a + 2√3 = 0,
II. 3b2– (6 + 2√3)b + 4√3 = 0

Ans:4
6a2 – (3 + 4√3)a + 2√3 = 0
(6a2 – 3a) – (4√3a – 2√3) = 0
3a (2a- 1) – 2√3 (2a – 1) = 0,
So a = 1/2, 2√3/3 (1.15)
3b2 – (6 + 2√3)b + 4√3 = 0
(3b2 – 6b) – (2√3b – 4√3) = 0
3b (b – 2) – 2√3 (b – 2) = 0
So, b = 2, 2√3/3a ≤ b

5.
I. 8a2 + (4 + 2√2)a + √2 = 0
II. b2 – (3 + √3)b + 3√3 = 0

Ans:2
8a2 + (4 + 2√2)a + √2 = 0
(8a2 + 4a) + (2√2a + √2) = 0
4a (2a + 1) + √2 (2a + 1) = 0
So a = -1/2 (-0.5), -√2/4 (-0.35)
b2 – (3 + √3)b + 3√3 = 0
(b2 – 3b) – (√3b – 3√3) = 0
b (b – 3) – √3 (b – 3) = 0
So b = 3, √3 (1.73)a < b

Directions:6-10) In each of the following questions, two equations are given. You have to solve these questions and find out the values of p and q.

6.
I.16p² + 20p + 6 = 0
II.10q² + 38q + 24 = 0

Ans:1
I. 16p² + 20p + 6 = 0
or, 8p² + 10p + 3 = 0
or, (4p + 3 ) ( 2p + 1) = 0
Therefore, p = -3/4 = -0.75 or p = – 1/2 = – 0.5II. 10q² + 38q + 24 = 0
or, 5q² + 19q + 12 = 0
or, ( q + 3) (5q + 4 ) = 0
Therefore, q = – 3 or q = – 4/5 = -0.8
Hence p > q

7.
I.18p² + 18p + 4 = 0
II.12q² + 29q + 14 = 0

Ans:1

  1. 18p² + 18p + 4 = 0
    or, 9p² + 9p + 2 = 0
    or, ( 3p + 2 ) ( 3p + 1) = 0
    Therefore,  p = – 2/3 = -0.67      or    p =  – 1/3 = -0.33
  2. 12q² + 29q + 14 = 0
    or, (3q + 2)  ( 4q + 7 ) = 0
    Therefore,   q = – 2/3 = -0.67      or       q = -7/4 = -1.75
    Hence  p ≥ q

8.
I.8p² + 6p = 5
II.12q² – 22q + 8 = 0

Ans:5
I. 8p² + 6p – 5 = 0
or, (4p + 5) (2p – 1) = 0
Therefore, p = – 5/4 = -1.25 or p = 1/2 = 0.5II. 12q² – 22q + 8 = 0
or, 6q² -11q + 4 = 0
or, (2q -1) (3q – 4) = 0
Therefore, q = 1/2 = 0.5 or q = 4/3 = 1.33
Hence p ≤ q

9.
I.17p² + 48p – 9 = 0
II.13q² = 32q – 12

Ans:3
I. 17p² + 48p – 9 = 0
or, (p + 3) (17p – 3) = 0
Therefore, p = -3 or p = 3/17 = 0.18II. 13q² – 32q + 12 = 0
or, (q -2) (13q – 6) = 0
Therefore, q = 2 or q = 6/13 = 0.46
Hence p < q

10.
I.821p² – 757p² = 256
II.√196 q³ – 12q³ = 16

Ans:1
64p² = 256
or, p² = 4
or, p = ±2
AND
14q³ – 12q³ = 16
or, 2q³ = 16
or, q³ = 8
or, q = 2
Hence p ≤ q

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