Download quick test quadratic equation for ibps

QUICK TEST 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 no relation can be established between ‘x’ and ‘y’.
1. I. x² + 3x - 28 = 0
II. y² – 11y + 28 = 0
2. I. 6x² – 17x + 12 = 0
II. 6y² – 7y + 2 = 0
256
576
3. I. x 
II. 3y² + y - 2 = 0
4. I. x² = 64
II. y² = 9y
5. I. x² + 6x - 7 = 0
II. 41y + 17 = 140
6. I. x2 + 3x = 28
II. y2 + 16y + 63 = 0
7. I. x =
3
2197
II. y2 = 169
8. I. 8x2 – 49x + 45 = 0
II. 8y2 – y – 9 = 0
9. I. 42x – 17y = –67
II. 7x + 12y = –26
10. I. x2 – 8x + 15 = 0
II. 2y2 – 21y + 55 = 0
ANSWERS WITH EXPLANATION
1. 4;
or,
or,

II.
or,
or,
or,

I. x² + 7x – 4x – 28 = 0
x(x + 7) – 4 (x + 7) = 0
(x – 4)(x + 7)=0
x = 4, – 7
y² – 11y + 28 = 0
y² – 7y – 4y + 28 = 0
y (y – 7) –4(y – 7) = 0
(y – 4) (y – 7) = 0
y = 4, 7
xy
2. 1; I. 6x² – 17x + 12 = 0
or, 6x² – 9x – 8x + 12 = 0
or, 3x (2x – 3) – 4 (2x – 3)=0
Downloaded From: http://bsc4success.com
or, (3x – 4) (2x – 3) = 0
x 
4 3
,
3 2
bsc
4su
cce
ss.c
om
II. 6y² – 3y – 4y + 2 = 0
or, 3y (2y – 1) – 2 (2y – 1) =0
or, (3y – 2) (2y – 1) = 0
2 1
,
3 2
 y=
 x>y
3. 2; I. x 
x 
II.
or,
or,
or,
256
576
16 2

24 3
3y² + y – 2 = 0
3y² + 3y – 2y – 2 = 0
3y (y + 1) – 2(y + 1) = 0
(3y – 2) (y + 1) = 0
2
 y=3,–1
 xy
4. 5; I. x² = 64
x =  8
II. y² = 9y
or, y² – 9y=0
or, y (y – 9) = 0
 y=0,9
 no relationship can be established between x and y.
5. 3; I. x² + 6x – 7 = 0
or, x² + 7x – x – 7 = 0
or, x(x + 7) –1 (x + 7)= 0
or, (x – 1) (x + 7) = 0
 x = 1, –7
II. 41y + 17 = 140
or, 41y = 140 – 17 = 123
 y=
123
3
41
 x<y
6. 2; I. x2 + 3x – 28 = 0
or, x2 + 7x – 4x – 28 = 0
or, x (x + 7) –4 (x + 7) = 0
or, ( x– 4) (x + 7) =0
or, x = 4, –7
II. y2 + 9y + 7y + 63 = 0
or, y (y + 9) +7 (y + 9) = 0
or, (y + 7) (y + 9) = 0
Downloaded From: http://bsc4success.com
or, y = –7, –9
xy
bsc
4su
cce
ss.c
om
7. 2; I. x = 3 2197
 x = 13
II. y2 = 169
 y = ±13
 xy
8. 2;
or,
or,
or,
II.
or,
or,
I. 8x2 – 40x – 9x + 45 = 0
8x (x – 5) –9 (x – 5) = 0
(8x – 9) (x – 5) = 0
x = 5, 9/8
8y2 + 8y – 9y –9 = 0
8y (y + 1) –9 (y + 1) = 0
(8y – 9) (y + 1) = 0
9
8
 y=
, –1
 xy
9. 3; 42x – 17y = – 67
42x + 72y = –156
–
–
+
eqn (II) × 6
–89y = 89
89
= –1 and x = –2
89

y=

x<y
10. 4; I. x2 – 8x + 15 = 0
or, x2 – 3x – 5x + 15 = 0
or, x (x – 3) –5 (x – 3) = 0
or, (x – 3) (x – 5) = 0
or, x = 3, 5
II. 2y2 – 10y + 55 = 0
or, 2y (y – 5) –11 (y – 5) = 0
or, (y – 5) (2y – 11) = 0
or, y = 5,

11
2
x y
Downloaded From: http://bsc4success.com