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

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We are providing the most important Quadratic Equations (Inequalities) for SBI PO 2019, SBI Clerk 2019 and all other competitive bank and insurance exams. These questions have very high chances to be asked in SBI PO 2019, SBI Clerk 2019.
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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

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

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

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

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

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

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

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

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

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

 

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

    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

    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

     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

    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

     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

    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

     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

    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

    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

    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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