Download Answers with Explanation.

Quadratic Equation for IBPS
bsc
4su
cce
ss.c
om
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 relationship between x and y cannot be established.
1
II. y  3 13824
II.13x – 17y + 128 = 0
II.y2 + 11y + 30 = 0
II.(y + 7)2 = 0
II.y2 – y – 6 = 0
II.y2 + 15y + 56 = 0
II.y2 + 13y + 42 = 0
II.y2 – 7y + 6 = 0
II.2y2 – 19y + 35 = 0
II.4y2 + 3y – 10 = 0
1. I.676x2 – 1 = 0
2. I.8x + 13y = 62
3. I.7x2 + 2x = 120
4. I.x2 = 7x
5. I.2x2 + 5x – 33 = 0
6. I.x2 + 12x + 36 = 0
7. I.x2 = 35
8. I.2x2 – 3x – 35 = 0
9. I.6x2 – 29x + 35 = 0
10. I.12x2 – 47x + 40 = 0
Answers with Explanation.
1. 3;
I.2x2 + 5x – 33 = 0
2
I.676x – 1 = 0
1
or, x 2  676
x  
1
y=
II. y  3 13824
II.y2 – y – 6 = 0
1
26
1
24
ie, x < y
2. 3; On solving these two equations, we get x = –2, y = 6
ie, x < y
3. 1;
I.7x2 – 28x + 30x – 120 = 0
or,7x(x – 4) +30(x – 4) = 0
or,(x – 4) (7x + 30) = 0
x = 4,

30
7
II.y2 + 6y + 5y + 30 = 0
or,y(y + 6) + 5(y + 6) = 0
or,(y + 5) (y + 6) = 0
y = –5, –6
ie, x > y
4. 1;
I.x2 = 7x
2
or,x – 7x = 0
or,x(x – 7) = 0
 x = 0, 7
Downloaded From: http://bsc4success.com
bsc
4su
cce
ss.c
om
II.(y + 7)2 = 0
or,(y + 7) = 0
y = –7
ie, x > y
5. 5;
I.2x2 – 6x + 11x – 33 = 0
or,2x(x – 3) +11(x – 3) = 0
or, (2x + 11) (x – 3) = 0
11
 x = 3,  2
II.y2 – 3y + 2y – 6 = 0
or,y(y – 3) +2(y – 3) = 0
or,(y + 2) (y – 3) = 0
y = –2, 3
ie, no relation exists between x and y.
6. 1;
I.x2 + 12x + 36 = 0
or,(x + 6)2 = 0
or,x + 6 = 0
or,x = – 6
II.y2 + 15y + 56 = 0
or,y2 +7y + 8y + 56 = 0
or,y(y + 7) +8(y + 7) = 0
or,(y + 7) (y + 8) = 0
 y = –7, –8
x>y
I.x2 = 35
7. 1;
 x   35
II.y2 + 13y + 42 = 0
or,y2 + 6y + 7y + 42 = 0
or,y(y + 6) +7(y + 6) = 0
or,(y + 6) (y + 7) = 0
y = –6, – 7
x > y
8. 5;
I.2x2 – 3x – 35 = 0
or,2x2 – 10x + 7x – 35 = 0
or,2x(x – 5) +7(x – 5) = 0
or, (x + 7) (x – 5) = 0
7
 x = 2 , 5
II. y2 – 7y + 6 = 0
or, y2 – y – 6y + 6 = 0
or, y(y – 1) – 6(y – 1)
or, (y – 1) (y – 6) = 0

y = 1, 6
No relation can be established between x and y.
Downloaded From: http://bsc4success.com
9. 4;
I.6x2 – 29x + 35 = 0
2
or,6x – 15x – 14x + 35 = 0
or,
3x(2x – 5) –7(2x – 5) = 0
or,(3x – 7) (2x – 5) = 0
7 5
,
3 2
bsc
4su
cce
ss.c
om

x=
2
II.2y – 19y + 35 = 0
or,2y2 – 14y – 5y + 35 = 0
or,2y(y – 7) –5(y – 7) = 0
or,(2y – 5) (y – 7) = 0
y =
5
,7
2
xy
10. 2;
I.12x2 – 47x + 40 = 0
2
or,12x – 32x – 15x + 40 = 0
or,4x(3x – 8) –5(3x – 8) = 0
or,(4x – 5) (3x – 8) = 0
5 8
x = 4 , 3
II.4y2 + 3y – 10 = 0
or,4y2 + 8y – 5y – 10 = 0
or,4y(y + 2) – 5(y + 2) = 0
or,(4y – 5) (y + 2) = 0
5
y = 4 ,  2
x  y
Downloaded From: http://bsc4success.com