QUADRATIC EQUATION QUESTIONS FOR IBPS SET 1

Below set has Quadratic questions for IBPS PO exam. As you all know 5 questions from quadratic equations are being asked in Bank PO.

Set 1 Quadratic equation questions for IBPS PO

 
In each of the following questions two equations are given. Solve these equations and give answer :
A) if x >= y, i.e x is greater than or equal to y
B) if x > y, i.e x is greater than y
C) if x <= y, i.e x is less than or equal to y
D) if x < y, i.e x less than y
E) if x = y,or no relation can be established between x and y

1.  I.x2 + 5 x + 6 = 0
   II.y2 +7 y + 12 = 0

2. I.x2 + 20 = 9x
   II.y2 + 42 = 13y

3. I.2x + 3y = 14
   II. 4x + 2y = 16

4. I.x = Sqrt625
    II.y = sqrt 676

5. I. x2 + 4x + 4 = 0
   II.y2 – 8y + 16 = 0

6. I.x2 – 19x + 84 = 0
    II.y2 – 25y + 156 = 0

7. I.x3 – 468 = 1729
    II.y2 – 1733 + 1564 = 0

8.  I. 9/√x + 19/√x = √x
   II.y5 – (2×14)11/2/√y = 0

9. I.√784x + 1234 = 1486
   II. √1089y + 2081 = 2345

10. I.12/√x – 23/√x = 5√x
   II.√y/12 -5√y/12 = 1/√y

ANSWERS with explanation

Answer 1. 
I.  x2 + 5x + 6 = 0
=>x2 + 2x + 3x + 6 = 0
=>x(x + 2) + 3(x + 2) = 0
=>(x + 3)(x + 2) =0
=> x= -3 or -2

II. y2+ 7y + 12 = 0
=>y2 + 4y + 3y + 12 = 0
=>y(y + 4) + 3 (y + 4) = 0
=>(y + 3) (y + 4) = 0
=> y = – 3 or – 4

Answer 2. 
I.  x2 – 9x + 20 = 0
x 2 – 5x – 4x + 20 = 0
x(x – 5) – 4 (x – 5) = 0
(x – 4) (x – 5) = 0
x = 4 or 5

II. y2 – 13 y + 42 = 0
=>y2 – 7y – 6y + 42 = 0
=>y(y – 7) – 6(y – 7) = 0
=>(y – 6) (y – 7) = 0
=> y = 6 or 7

Answer 3. 2x + 3y = 14….(i)
   4x + 2y = 16….(ii)

On solving, (i) x 2 – (ii) we get,

 4x + 6y – 4x – 2y = 28 – 16
 => 4y = 12 =>y = 3
 So, the value of x will be –
 2x + 3 x 3 = 14
 =>2x = 14 – 9 = 5 => x = 5/2

Answer 4. 
I.  x = √625 = +25 or -25
II. y = √676 = +26 or -26

Answer 5. 
I.  x 2 + 4x + 4 = 0
(x + 2)2 = 0 => x = – 2

II.  y2 – 8y + 16 = 0
=>(y – 4)2 = 0
=>y = 4

Answer 6. 
I.  x2 – 19 x + 84 = 0
=> x2 – 12 x – 7 x + 84 = 0
=> x(x – 12) – 7 (x – 12) = 0
=> (x – 7) (x – 12) = 0
So, x = 7 or x = 12

II.  y2 – 25 y + 156 = 0
=> y2 – 12 y – 13 y + 156 = 0
=> y(y – 12) – 13 (y – 12) = 0
=> (y – 13) (y – 12) = 0
So, y = 12 or 13

Answer 7. 
I. x3 = 1729 + 468 = 2197
So, x = 3√2197 = 13

II. y2 = 1733 – 1564 = 169
So, y = √169 = + 13 or – 13

Answer 8. 
I. 9 + 19 = √x + √x
=> 28 = x

II.  y 5 x y1/2 = (28)11/2
=> y11/2 = (28)11/2
=> y = 28

Answer 9. I. √784 x = 1486 – 1234 = 252
=> x = 252/√784 = 252/28 = 9

II. √1089y = 2345 – 2081 = 264
=>y = 264/√1089 = 264/33 = 8

Answer 10. 
I.  12 – 23 = 5 √x x √x
=>-11 = 5x
=>x = -11/5

II.  (√y – 5√y)/12 = 1/√y
=> -4√y x √y = 12
=> -4y = 12
=> y = -3

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  • Nirmal Agarwal

    4th one is wrong actually its y>x

    • Hi Nirmal,

      The answer is correct

      • Nirmal Agarwal

        if
        x^2=9
        x=+-3
        and
        x=root 9
        x=3

        considering this example 4th one is wrong ??