Quadratic Equation for SBI PO Questions

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.

Quadratic Equation for SBI PO Prelims Questions

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

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

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

So, x ≤ y

Hence, option e.

Quadratic Equation for SBI PO Mains Questions

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

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

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

So, the relationship cannot be established

Hence, option c.

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