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Quadratic Equation for LIC AAO/IBPS
Directions (Q. 1-10): In each of these questions, two equations (I) and (II) are given.
You have to solve both the equations and give answer
1) if x > y
2) if x > y
3) if x < y
4) if x < y
5) if x = y or no relation can be established between x and y.
1.
I. x2 – 8 3 x + 45 = 0
II. y2 –
2.
I. x – 7 2x + 24 = 0
II. y – 5 2 y + 12 = 0
3.
I. 12x2 – 17x + 6 = 0
II. 20y2 – 31y + 12 = 0
4.
I. 3x2 – 8x + 4 = 0
II. 4y2 – 15y + 9 = 0
5.
I. x2 – 16x + 63 = 0
II. y2 – 2y – 35 = 0
6.
I. 4x + 7y = 42
II. 3x – 11y = –1
7.
I. 9x2 – 29x + 22 = 0
II. y2 – 7y + 12 = 0
8.
I. 3x2 – 4x – 32 = 0
II. 2y2 – 17y + 36 = 0
9.
I. 3x2 – 19x – 14 = 0
II. 2y2 + 5y + 3 = 0
10. I. x2 + 14x + 49 = 0
2 y – 24 = 0
II. y2 + 9y = 0
Answers with Explanation
1. 5; I. x2 – 8 3x + 45 = 0
or x2 – 5 3x + 3 3x + 45 = 0
or x(x – 5 3 ) + 3 3 (x – 5 3 ) = 0
or (x + 3 3 ) (x – 5 3 ) = 0
x= 3 3, 5 3
II. y2 –
2 y – 24 = 0
or y2 – 4 2 y + 3 2 y – 24 = 0
or (y – 4 2 y ) (y + 3 2 ) = 0
or (y + 3 2 ) (y – 4 2 )

y = –3 2 , 4 2
Hence relation cannot be established between x and y.
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2. 2; I. x – 7 2x + 24 = 0
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or x – 4 2x – 3 2x + 24 = 0
or
x ( x – 4 2) – 3 2( x – 4 2)  0
or ( x – 3 2 )( x – 4 2 )  0
Now, if
x –3 2 0
then x = 3 2
 x = 9 × 2 = 18
If x – 4 2 = 0
then x  4 2
 x = 16 × 2 = 32
II. y – 5 2 y + 12 = 0
or y – 3 2 y – 2 2 y + 12 = 0
or
y( y – 3 2) – 2 2( y – 3 2)  0
or ( y – 2 2 ) ( y – 3 2 ) = 0
If ( y – 2 2 ) = 0
then
y =2 2
y=4×2=8
If
y – 3 2 = 0
then
y 3 2
 y = 9 × 2 = 18
xy
3. 4; I. 12x2 – 17x + 6 = 0
or 12x2 – 9x – 8x + 6 = 0
or 3x(4x – 3) – 2(4x – 3) = 0
or (3x – 2) (4x – 3) = 0
If 3x – 2 = 0
then 3x = 2
2
3
If 4x – 3 = 0
x=
3
4
2
II. 20y – 31y + 12 = 0
or 20y2 – 16y – 15y + 12 = 0
or 4y(5y – 4) – 3(5y – 4) = 0
or (4y – 3) (5y – 4) = 0
then x =
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3 4
,
4 5
Hence x < y
4. 5; I. 3x2 – 8x + 4 = 0
or 3x2 – 6x – 2x + 4 = 0
or (3x – 2) (x – 2) = 0
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y=
2
3
2
II. 4y – 15y + 9 = 0
or 4y2 – 12y – 3y + 9 = 0
or 4y(y – 3) – 3(y – 3) = 0
or (4y – 3) (y – 3) = 0
 x = 2,
3
,3
4
Relation cannot be established between x and y.
5. 1; I. x2 – 16x + 63 = 0
or x2 – 9x – 7x + 63 = 0
or x(x – 9) – 7(x – 9) = 0
or (x – 7) (x – 9) = 0
 x = 7, 9
II. y2 – 2y – 35 = 0
or y2 – 17y + 5y – 35 = 0
or y(y – 7) + 5(y – 7) = 0
or (y + 5) (y – 7) = 0
 y = –5, 7
Hence, x > y
y=
6. 1; eqn (I) ×3 – eqn (II) × 4
12x  21y  126
12x – 44 y  – 4
– 

65y  130
y = 2
andx = 7
7. 3; I. 9x2 – 18x – 11x + 22 = 0
or 9x(x – 2) – 11(x – 2) = 0
or (x – 2) (9x – 11) = 0
11
9
II. y2 – 3y – 4y + 12 = 0
or y(y – 3) – 4(y – 3) = 0
or (y – 3) (y – 4) = 0
y = 3, 4
x < y
x = 2,
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8. 4; I. 3x2 – 4x – 32 = 0
or 3x2 – 12x + 8x – 32 = 0
or 3x(x – 4) + 8(x – 4) = 0
or (3x + 8) (x – 4) = 0
8
3
2
II. 2y – 8y – 9y + 36 = 0
or 2y(y – 4) – 9(y – 4) = 0
or (2y – 9) (y – 4) = 0
or (2y – 9) (y – 4) = 0
x = 4, –
y = 4,
9
2
x < y
9. 1; I. 3x2 – 21x + 2x – 14 = 0
or 3x(x – 7) + 2(x – 7) = 0
or (3x + 2) (x – 7) = 0
2
3
2
II. 2y + 5y + 3 = 0
or 2y2 + 2y + 3y + 3 = 0
or 2y(y + 1) + 3(y + 1) = 0
or (2y + 3) (y + 1) = 0
x = 7, –
y = –
3
, –1
2
x > y
10. 5; I. x2 + 14x + 49 = 0
or (x + 7)2 = 0
x+7=0
or,x = –7
II. y2 + 9y = 0
or y(y + 9) = 0
y = 0, –9
ie no relation between x and y.
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