Set-24 Quadratic Equations (Inequalities) For SBI PO and SBI Clerk 2019 | Must Go Through These Questions

• p ≤ q
• p > q
• Relation cannot be established or p = q
• p < q
• p ≥ q

Check your Answers below:

• 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. Question

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

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  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. Question

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

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  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. Question

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

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
    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. Question

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

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
  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. Question

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

  • If a > b
  • If a < b
  • If a ≥ b
  • If a ≤ b
  • If a = b or relation cannot be established
    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. Question

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

  • p > q
  • p ≤ q
  • Relation cannot be established or p = q
  • p ≥ q
  • p < q
  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. Question

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

  • p ≥ q
  • p > q
  • p < q
  • p ≤ q
  • Relation cannot be established or p = q

    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. Question

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

  • p < q
  • Relation cannot be established or p = q
  • p ≥ q
  • p > q
  • p ≤ q
  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. Question

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

  • Relation cannot be established or p = q
  • p > q
  • p < q
  • p ≥ q
  • p ≤ q
  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. Question

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

  • p ≤ q
  • p > q
  • Relation cannot be established or p = q
  • p < q
  • p ≥ q
  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