Quadratic Equation: Formula to Find Roots & Example Problems

Quadratic Equation:

Quadratic Equation is one of the most important sections of quantitative aptitude that is asked in the Banking exams like SBI Clerk, IBPS Clerk, SBI PO, IBPS PO, etc. Every year in the banking preliminary examination there are at least 4 to 5 questions that are asked from quadratic equation topic. One of the most important benefits of the quadratic equation problems is that they are easy and can easily fetch at least 3 4 marks in less than 3 minutes. However, the candidates picking the quadratic equation questions should be well versed in the ways to solve them. If the candidates know the correct way to calculate the roots of the quadratic equation then it can prove to be a game-changer for the candidates. Here we have given some quadratic equation examples. You can refer to them and learn how to solve quadratic equation problems. You can also check how to calculate the sum of roots of quadratic equation.

With this in mind, we have come up with this article on major quadratic equation examples and the ways to solve them.

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What is a Quadratic Equation?

The Quadratic Equation is one of the branches of Algebra, wherein X represents the unknown variables while a,b, and c represents constants or numbers arranged in the form i.e. ax²+ bx+c = 0. Once the candidate can derive the roots of the quadratic equation then he can easily solve the question. The candidates can use the following formula to derive the sum of roots of the quadratic equation:

(α, β) = [-b ± √(b² – 4ac)]/2ac

Although the above formula is not widely used yet the candidates should first try to know what does roots of the quadratic equation usually means:

Sum of Roots of Quadratic Equation

The roots of any quadratic equation are nothing but the values that satisfies the whole equation. For example,

In the equation, x² – x – 20 = 0 upon solving the equation the quadratic equation, we get the following answer

x² – x – 20 = 0

x² – 5x + 4x – 20 = 0

x(x – 5) + 4(x – 5) = 0

(x + 4)(x – 5) = 0

x = -4, 5

This implies that -4 and 5 are the roots of the quadratic equation, which means if you put the values -4 or 5 in the place of X in the equation above the value will come down to 0 thus satisfying the LHS=RHS thing.

How to Solve Quadratic Equation?

The candidates can follow the quadratic equation question with the answer given in the space below to get an idea of the best process on how to solve quadratic equation:

1.x² – x – 20 = 0

y²– 9y – 22 = 0

A. x > y

B. x ≥ y

C. x = y or relationship can’t be determined.

D. x < y

E. x ≤ y

Ans:

x² – x – 20 = 0

x² – 5x + 4x – 20 = 0

x(x – 5) + 4(x – 5) = 0

(x + 4)(x – 5) = 0

x = -4, 5

y² – 9y – 22 = 0

y² – 11y + 2y – 22 = 0

y(y – 11) + 2(y – 11) = 0

(y + 2)(y – 11) = 0

y = -2, 11

The relationship between x and y cannot be established.

2.5x² + 10 = 55

y²– 8y + 15 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Ans. 5x² + 10 = 55

5x² = 45

x = 3, -3

y² – 8y + 15 = 0

y² – 5y – 3y + 15 = 0

y(y – 5) – 3(y – 5) = 0

(y – 3)(y – 5) = 0

y = 3, 5

x ≤ y

x² – 7x + 12 = 0

y²– 3y + 2 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Ans.

x² – 7x + 12 = 0

x² – 3x – 4x + 12 = 0

x(x – 3) – 4(x – 3) = 0

(x – 4)(x – 3) = 0

x = 4, 3

y² – 3y + 2 = 0

y² – 2y – y + 2 = 0

y(y – 2) – 1(y – 2) = 0

(y – 1)(y – 2) = 0

y = 1, 2

x > y

2x + 4y = 68

5x – 2y = 32

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Ans.

2x + 4y = 68 —-(1)

5x – 2y = 32 —-(2)

12x = 132

x = 11

y = 11.5

x < y

√x = 3

y²= 16

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Ans.

√x = 3

x = 9

y² = 16

y = 4, -4

x > y

Top 50 Quadratic Equation PDF Free Download

Previous Year Quadratic Equation Questions with Answers

The Quadratic Equations have been asked over the years in the banking exams. In other words, candidates who have gained mastery over this topic find it very easy to fetch around 4-5 marks in no time. Thus, Quadratic Equations is one of the best and the easiest way to score more marks in the Quantitative Aptitude section.

1.x²+ 9x + 20 = 0

y²+ 13y + 42 = 0

x > y
x ≥ y
x = y or relationship can’t be determined.
x < y
x ≤ y

Ans.

x² + 9x + 20 = 0

x² + 4x + 5x + 20 = 0

x(x + 4) + 5(x + 4) = 0

(x + 5)(x + 4) = 0

x = -5, -4

y² + 13y + 42 = 0

y² + 6y + 7y + 42 = 0

y(y + 6) + 7(y + 6) = 0

(y + 7)(y + 6) = 0

y = -7, -6

x > y

x²– 18x + 72 = 0

y²– 21y + 108 = 0

x > y
x ≥ y
x = y or relationship can’t be determined.
x < y
x ≤ y

Ans.

x² – 18x + 72 = 0

x² – 12x – 6x + 72 = 0

x(x – 12) – 6(x – 12) = 0

x = 6, 12

y² – 21y + 108 = 0

y² – 12y – 9y + 108 = 0

y(y – 12) – 9(y – 12) = 0

(y – 9)(y – 12) = 0

y = 9, 12

The relationship between x and y cannot be established.

2x²– 2x – 84 = 0

y²– 2y – 48 = 0

x > y
x ≥ y
x = y or relationship can’t be determined.
x < y
x ≤ y

Ans.

2x² – 2x – 84 = 0

2x² – 14x + 12x – 84 = 0

2x(x – 7) + 12(x – 7) = 0

(2x + 12)(x – 7) = 0

x = -6, 7

y² – 2y – 48 = 0

y² – 8y + 6y – 48 = 0

y(y – 8) + 6(y – 8) = 0

(y + 6)(y – 8) = 0

y = -6, 8

The relationship between x and y cannot be established.

2x + y = 34

x + 3y = 47

x > y
x ≥ y
x = y or relationship can’t be determined.
x < y
x ≤ y

Ans.

2x + y = 34 ——–(1)

x + 3y = 47 —–(2)

(1) – (2) * 2

-5y = -60

y = 12

x = 47 – 36 = 11

x < y

x²+ 32x – 144 = 0

y²– 31y + 108 = 0

x > y
x ≥ y
x = y or relationship can’t be determined.
x < y
x ≤ y

Ans.

x² + 32x – 144 = 0

x² + 36x – 4x – 144 = 0

x(x + 36) – 4(x + 36) = 0

(x – 4)(x + 36) = 0

x = 4, -36

y² – 31y + 108 = 0

y² – 27y – 4y + 108 = 0

y(y – 27) – 4(y – 27) = 0

(y – 4)(y – 27) = 0

y = 4, 27

x ≤ y

Important Quadratic Equation Examples

The following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give an answer as,

a) If x > y
b) If x ≥ y
c) If x = y or relationship can’t be determined.
d) If x < y
e) If x ≤ y

1.x²+ 2x – 195 = 0

y²+ y – 182 = 0

Ans. x² + 2x – 195 = 0

x² + 15x – 13x – 195 = 0

x(x + 15) – 13(x + 15) = 0

(x – 13)(x + 15) = 0

X = 13, -15

y² + y – 182 = 0

y² + 14y – 13y – 182 = 0

y(y + 14) – 13(y + 14) = 0

(y – 13)(y + 14) = 0

Y = 13, -14

The relationship between x and y cannot be established.

2.x²– 25x + 136 = 0

y²– y – 56 = 0

Ans. x² – 25x + 136 = 0

x² – 17x – 8x + 136 = 0

x(x – 17) – 8(x – 17) = 0

(x – 8)(x – 17) = 0

x = 8, 17

y² – y – 56 = 0

y² – 8y + 7y – 56 = 0

y(y – 8) + 7(y – 8) = 0

(y + 7)(y – 8) = 0

y = -7, 8

Hence, x ≥ y

3.2x²+ 16x + 30 = 0

y²+ 6y + 8 = 0

Ans. 2x² + 16x + 30 = 0

2x² + 10x + 6x + 30 = 0

2x(x + 5) + 6(x + 5) = 0

(2x + 6)(x + 5) = 0

x = -3, -5

y² + 6y + 8 = 0

y² + 4y + 2y + 8 = 0

y(y + 4) + 2(y + 4) = 0

(y + 2)(y + 4) = 0

y = -2, -4

The relationship between x and y cannot be established.

x2+ 5x + 6 = 0

y2– 2y + 1 = 0

Ans. x² + 5x + 6 = 0

x² + 3x + 2x + 6 = 0

x(x + 3) + 2(x + 3) = 0

(x + 2)(x + 3) = 0

x = -2, -3

y² – 2y + 1 = 0

y² – y – y + 1 = 0

y(y – 1) – 1(y – 1) = 0

(y – 1)(y – 1) = 0

y = 1, 1

Hence, x < Y

x²– 11x + 30 = 0

y²– 21y + 90 = 0

Ans. x² – 11x + 30 = 0

x² – 6x – 5x + 30 = 0

x(x – 6) – 5(x – 6) = 0

(x – 5)(x – 6) = 0

x = 5, 6

y² – 21y + 90 = 0

y² – 15y – 6y + 90 = 0

y(y – 15) – 6(y – 15) = 0

(y – 6)(y – 15) = 0

y = 6, 15

Hence, x ≤ y

Practice More Quadratic Equation

Quadratic Equation Quiz 1

Quadratic Equation Quiz 2

Quadratic Equation Quiz 3

Quadratic Equation Quiz 4

More 1000+ Quadratic Equation Practice Questions

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Quadratic Equation: Formula to Find Roots & Example Problems