Quadratic Equation For IBPS RRB Exam, Check Important Question

Quadratic Equation for IBPS RRB Exam- The topic of quadratic equations is a crucial part of the quantitative aptitude section in various competitive exams, including the IBPS RRB exam and other banking exams. Here we are going to discuss a few Quant Quadratic Equation Practice Questions for the Prelims Exam. These questions will be helpful for the upcoming IBPS RRB 2024 Exam. Though it is one of the trickiest topics from this section, these can be solved with constant practice which is why we are here with these practice questions along with their answers and detailed solutions. Mastering these equations is essential for scoring well in this section. Candidates check all details in the article below.

What is a Quadratic Equation?

A quadratic equation is a second-degree polynomial equation in one variable of the form:

ax² + bx + c = 0

where a, b and c are constants, and the x is variable. The x represents the unknown quantity, and the constants a, b and c determine the shape and nature of the equation.

Important Question for Quadratic Equation

Question 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. 3x² – 24x + 36 = 0

II. 4y² = y + 5

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 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 – 2x – 63 = 0

II. y² + 14y + 48 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 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. x² + 15x + 54 = 0

II. y² + 20y + 99 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 4: 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. 3x + 7y = 53

II. 7x – 5y = 17

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 5: 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 – 27x + 180 = 0

II. y² – 24y + 140 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 6: 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 – 11x + 30 = 0

II. y² – 11y + 18 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 7: 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 + 3x – 10 = 0

II. y² – 5y + 6 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 8: 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 – 5x + 4 = 0

II. y² + 2y – 3 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 9: 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. 3x + 5y = 21

II. 9x – 2y = 12

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 10: 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. 4x² + 16x – 20 = 0

II. 2y² + 11y – 6 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Solution 1: A)

From I:

3x² – 24x + 36 = 0

3x² – 18x – 6x + 36 = 0

3x(x – 6) – 6(x – 6) = 0

(3x – 6)(x – 6) = 0

x = 2, 6

From II:

4y² = y + 5

4y² – y – 5 = 0

4y² – 5y + 4y – 5 = 0

y(4y – 5) + 1(4y – 5) = 0

(4y – 5)(y + 1) = 0

y = -1, 5/4

So x > y

Hence, option a.

Solution 2: C)

From I:

x² – 2x – 63 = 0

x² – 9x + 7x – 63 = 0

x(x – 9) + 7(x – 9) = 0

(x + 7)(x – 9) = 0

x = 9, -7

From II:

y² + 14y + 48 = 0

y² + 8y + 6y + 48 = 0

y(y + 8) + 6(y + 8) = 0

(y + 8)(y + 6) = 0

y = -6, -8

No relation can be established between x and y.

Hence, option c.

Solution 3: D)

From I:

x² + 15x + 54 = 0

x² + 9x + 6x + 54 = 0

x(x + 9) + 6(x + 9) = 0

(x + 6)(x + 9) = 0

x = -6, -9

From II:

y² + 20y + 99 = 0

y² + 11y + 9y + 99 = 0

y(y + 11) + 9(y + 11) = 0

(y + 11)(y + 9) = 0

y = -9, -11

So x ≥ y

Hence, option d.

Solution 4: A)

From I:

3x + 7y = 53

3x = 53 – 7y

x = (53 – 7y)/3

From II:

7x – 5y = 17

7 × {(53 – 7y)/3} – 5y = 17

371 – 49y – 15y = 51

64y = 320

y = 5

Now x = {53 – 7 × 5}/3

x = 18/3

x = 6

So x > y

Hence, option a.

Solution 5: C)

From I:

x² – 27x + 180 = 0

x² – 15x  – 12x + 180 = 0

x(x – 15) – 12(x – 15) = 0

(x – 15)(x – 12) = 0

x = 15, 12

From II:

y² – 24y + 140 = 0

y² – 14y – 10y + 140 = 0

y(y – 14) – 10(y – 14) = 0

(y – 10)(y – 14) = 0

y = 14, 10

No relation can be established between x and y.

Hence, option c.

Solution 6: C)

From I:

x² – 11x + 30 = 0

x² – 6x – 5x + 30 = 0

x(x – 6) – 5(x – 6) = 0

(x – 6)(x – 5) = 0

x = 6, 5

From II:

y² – 11y + 18 = 0

y² – 2y – 9y + 18 = 0

y(y – 2) – 9(y – 2) = 0

(y – 2)(y – 9) = 0

y = 2, 9

Hence, option c.

Solution 7: E)

From I:

x² + 3x – 10 = 0

x² – 2x + 5x – 10 = 0

x(x – 2) + 5(x – 2) = 0

(x – 2)(x + 5) = 0

x = 2, -5

From II:

y² – 5y + 6 = 0

y² – 3y – 2y + 6 = 0                        

y(y – 3) – 2(y – 3) = 0

(y – 3)(y – 2) = 0

y = 3, 2

So, x ≤ y

Hence, option e.

Solution 8: D)

From I:

x² – 5x + 4 = 0

x² – x – 4x + 4 = 0

x(x – 1) – 4(x – 1) = 0

(x – 1)(x – 4) = 0

x = 1, 4

From II:

y² + 2y – 3 = 0

y² + 3y – y – 3 = 0

y(y + 3) – 1(y + 3) = 0

(y + 3)(y – 1) = 0

y = -3, 1

Hence, option d.

Solution 9: B)

From I:

3x + 5y = 21

3x = 21 – 5y

x = (21 – 5y)/3

From II:

9x – 2y = 12

9 × {(21 – 5y)/3} – 2y = 12

(189 – 45y – 6y)/3 = 12              

189 – 51y = 36

51y = 153, y = 3

x = {(21 – 5y)/3} = 6/3 = 2

So, x < y

Hence, option b.

Solution 10: C)

From I:

4x² + 16x – 20 = 0

4x² + 20x – 4x – 20 = 0

x (4x + 20) – 1(4x + 20) = 0

(4x + 20)(x – 1) = 0

x = -5, 1

From II:

2y² + 11y – 6 = 0

2y² + 12y – y – 6 = 0

2y(y + 6) – 1(y + 6) = 0

(y + 6)(2y – 1) = 0

y = – 6, 1/2

Hence, option c.

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