Quadratic Equation for SBI PO: The questions from the quadratic equation are more or less tricky and difficult in nature. But by practicing the right set of questions they are as easy as any other quant topic. We today have six questions, 3 each of prelims and mains with answer key.
1.) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. (x + 4)2 = 23x – 20
II. (y – 3)2 = 3 – y
a.) x > y
b.) x < y
c.) x = y or the relationship cannot be established
d.) x ≥ y
e.) x ≤ y
Answer: d)
Solution:
From I:
(x + 4)2 = 23x – 20
x2 + 8x + 16 = 23x – 20
x2 – 15x + 36 = 0
x2 – 3x – 12x + 36 = 0
x(x – 3) – 12(x – 3) = 0
(x – 3)(x – 12) = 0
x = 12, 3
From II:
(y – 3)2 = 3 – y
y2 – 6y + 9 = 3 – y
y2 – 5y + 6 = 0
y2 – 2y – 3y + 6 = 0
y(y – 2) – 3(y – 2) = 0
(y – 2)(y – 3) = 0
y = 2, 3
| X | Relation | y |
| 12 | > | 2 |
| 12 | > | 3 |
| 3 | > | 2 |
| 3 | = | 3 |
So, x ≥ y
Hence, option d.
2.) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 = 324
II. y2 + 18y = 0
a.) x > y
b.) x < y
c.) x = y or the relationship cannot be established
d.) x ≥ y
e.) x ≤ y
Answer: c)
Solution:
From I:
x2 = 324
x = -18, 18
From II:
y2 + 18y = 0
y(y + 18) = 0
y = 0, -18
| X | Relation | y |
| -18 | < | 0 |
| -18 | = | -18 |
| 18 | > | 0 |
| 18 | > | -18 |
So, the relationship cannot be established.
Hence, option c.
3.) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x3 = -729
II. y2 + 15y + 54 = 0
a.) x > y
b.) x < y
c.) x = y or the relationship cannot be established
d.) x ≥ y
e.) x ≤ y
Answer: e)
Solution:
From I:
x3 = -729
x = -9
From II:
y2 + 15y + 54 = 0
y2 + 6y + 9y + 54 = 0
y(y + 6) + 9(y + 6) = 0
(y + 9)(y + 6) = 0
y = -9, -6
| X | Relation | y |
| -9 | = | -9 |
| -9 | < | -6 |
So, x ≤ y
Hence, option e.
1.) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x + 19√x + 90 = 0
II. y + 23√y + 130 = 0
a.) x > y
b.) x < y
c.) x = y or the relationship cannot be established
d.) x ≥ y
e.) x ≤ y
Answer: e)
Solution:
From I:
x + 19√x + 90 = 0
x + 9√x + 10√x + 90 = 0
√x(√x + 9) + 10(√x + 9) = 0
(√x + 10)(√x + 9) = 0
√x = -10, -9
x = 100, 81
From II:
y + 23√y + 130 = 0
y + 10√y + 13√y + 130 = 0
√y(√y + 10) + 13(√y + 10) = 0
(√y + 10)(√y + 13) = 0
√y = -10, -13
y = 100, 169
| x | Relation | y |
| 100 | = | 100 |
| 100 | < | 169 |
| 81 | < | 100 |
| 81 | < | 169 |
So, x ≤ y
Hence, option e.
2.) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. x2 + (9√3)x + 60 = 0
II. y2 + (6 + 4√3)y + 24√3 = 0
a.) x > y
b.) x < y
c.) x = y or the relationship cannot be established
d.) x ≥ y
e.) x ≤ y
Answer: e)
Solution:
From I:
x2 + (9√3)x + 60 = 0
x2 + (5√3)x + (4√3)x + 60 = 0
x(x + 5√3) + 4√3(x + 5√3) = 0
(x + 5√3)(x + 4√3) = 0
x = -5√3, -4√3
From II:
y2 + (6 + 4√3)y + 24√3 = 0
y2 + 6y + (4√3)y + 24√3 = 0
y(y + 6) + 4√3(y + 6) = 0
(y + 4√3)(y + 6) = 0
y = -4√3, -6
| x | Relation | y |
| -5√3 | < | -4√3 |
| -5√3 | < | -6 |
| -4√3 | = | -4√3 |
| -4√3 | < | -6 |
So, x ≤ y.
Hence, option e.
3.) In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.
I. 81x2+ 144x + 55 = 0
II. 81y2+ 180y + 91 = 0
a.) x > y
b.) x < y
c.) x = y or the relationship cannot be established
d.) x ≥ y
e.) x ≤ y
Answer: c)
Solution:
From I:
81x2+ 144x + 55 = 0
81x2+ 99x + 45x + 55 = 0
9x(9x + 11) + 5(9x + 11) = 0
(9x + 5)(9x + 11) = 0
x = -5/9, -11/9
From II:
81y2+ 180y + 91 = 0
81y2+ 117y + 63y + 91 = 0
9y(9y + 13) + 7(9y + 13) = 0
(9y + 7)(9y + 13) = 0
y = -7/9, -13/9
| x | Relation | y |
| -5/9 | > | -7/9 |
| -5/9 | > | -13/9 |
| -11/9 | < | -7/9 |
| -11/9 | > | -13/9 |
So, the relationship cannot be established
Hence, option c.
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