Set-21 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 I and II are given. You have to solve both the equations.
Give Answer

1.
I. 15a2 – 10a – 5 = 0;
II. 20b2 + 18b + 4 = 0

2.
I. 26a2 – 88 = 16;
II. 3√b/8 – √b/4 = 1/√b

3.
I. 2a + 3b = 20;
II. 4a + 2b = 24

4.
I. 10a2 + 9a -7 = 0;
II. 12b + 84 = 122

5.
I. a2 = 196;
II. b2 + 11b + 24 = 0

Directions:6-10) In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.

6.
i) 2x2+11x+12=0
ii) x2+2x-35=0

7.
i) x2-20x+99=0
ii) 2y2-15y+28=0

8.
i) 2x+5y-√2601=0
ii) 7x+3y-77=0

9.
i) x2+19y+90=0
ii) 14y2-37y+24=0

10.
i) 5x2+36√3x+192=0
ii) 2y2+9√3y+27=0

 

Check your Answers below:

 

 

  • Directions:1-5) In the following questions, two equations I and II are given. You have to solve both the equations.

    Give Answer

    1. Question

    I. 15a2 – 10a – 5 = 0;
    II. 20b2 + 18b + 4 = 0

    Ans:5
    15 a2 + 15a – 5a –  5 = 0

    15a (a + 1) – 5 (a + 1) = 0

    (15a – 5) (a + 1) = 0

    a = 1/3, -1

    20b2 + 10b + 8b + 4 = 0

    10b (2b +1) + 4 (2b + 1) = 0

    (10b + 4) (2b + 1) = 0

    b = -2/5,    -1/2

    Hence there is no relation.

  • 2. Question

    I. 26a2 – 88 = 16;
    II. 3√b/8 – √b/4 = 1/√b

    Ans:3
    26a2 = 16 + 88 = 104

    a2 = 104/26 = 4

    a = ±2

    3√b/8 – 2√b/8 = 1/√b

    √b/8 = 1/√b

    b = 8

    a < b.

  • 3. Question

    I. 2a + 3b = 20;
    II. 4a + 2b = 24

    Ans:5
    4a + 6b = 40 —-I

    4a + 2b = 24 —-II

    Solve I and II we get

    4b = 16

    b = 4

    Apply the b = 4 in eqn I

    a = 4

    a= b.

  • 4. Question

    I. 10a2 + 9a -7 = 0;
    II. 12b + 84 = 122

    Ans:3
    10a2 + 14a – 5a – 7 = 0

    2a (5a + 7) – 1 (5a + 7) = 0

    (2a – 1) (5a + 7) = 0

    a = 1/2, -7/5

    12b = 144 – 84 = 60

    b = 60/12 = 5

    a < b.

  • 5. Question

    I. a2 = 196;
    II. b2 + 11b + 24 = 0

    Ans:5
    a = ±14

    b2 + 8b + 3b + 24 = 0

    b (b + 8) + 3 (b + 8) = 0

    (b + 3) (b + 8) = 0

    b = -3, -8

    Hence there is no relation.

    Directions:6-10) In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.

    6. Question

    i) 2x2+11x+12=0
    ii) x2+2x-35=0

    Ans:5
    a) 2x2+11x+12=0

    2x2+8x+3x+12=0

    (x+4)(2x+3)=0

    x= -4 and -3/2

    b) x2+2x-35=0

    x2+7x-5x-35=0

    (x+7)(x-5)=0

    x=-7 or 5

    Hence the relationship cannot be determined

  • 7. Question

    i) x2-20x+99=0
    ii) 2y2-15y+28=0

    Ans:1
    a) x2-20x+99=0

    x2-9x-11x+99=0

    (x-9)(x-11)=0

    x=9 and 11

    b) 2y2-15y+28=0

    2y2-8y-7y+28=0

    (y-4)(2y-7)=0

    y= 7/2 and 4

    Thus x>y

  • 8. Question

    i) 2x+5y-√2601=0
    ii) 7x+3y-77=0

    Ans:1
    After solving above equations we have x= 8 and y=7

    Hence x>y

  • 9. Question

    i) x2+19y+90=0
    ii) 14y2-37y+24=0

    Ans:2
    a) x2+19y+90=0

    x2+9y +10y+90=0

    (x+9)(x+10)=0

    x= -9 and -10

    b) 14y2-37y+24=0

    14y2-21y-16y+24=0

    7y(2y-3)-8(2y-3)=0

    y= 3/2 or 8/7

    Hence x<y

  • 10. Question

    i) 5x2+36√3x+192=0
    ii) 2y2+9√3y+27=0

    Ans:2
    a)5x2+36√3x+192=0

    5x2+20√3x+16√3x+192=0

    (5x+16√3)(x+4√3)=0

    x=-4√3 or -16√3/5

    b) 2y2+9√3y+27=0

    2y2+6√3y+3√3y +27=0

    (2y+3√3)(y+3√3)=0

    y= -3√3 or -3√3/2

    Thus x<y

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