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## 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.

**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. ax^{2}+ 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 ± √(b^{2} – 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, x^{2} – x – 20 = 0 upon solving the equation the quadratic equation, we get the following answer

x^{2} – x – 20 = 0

x^{2} – 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.**x**^{2} – x – 20 = 0

**y ^{2}– 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: **

x^{2} – x – 20 = 0

x^{2} – 5x + 4x – 20 = 0

x(x – 5) + 4(x – 5) = 0

(x + 4)(x – 5) = 0

x = -4, 5

y^{2} – 9y – 22 = 0

y^{2} – 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.**5x**^{2} + 10 = 55

**y ^{2}– 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.** 5x^{2} + 10 = 55

5x^{2} = 45

x = 3, -3

y^{2} – 8y + 15 = 0

y^{2} – 5y – 3y + 15 = 0

y(y – 5) – 3(y – 5) = 0

(y – 3)(y – 5) = 0

y = 3, 5

x ≤ y

**x**^{2}– 7x + 12 = 0

**y ^{2}– 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. **

x^{2} – 7x + 12 = 0

x^{2} – 3x – 4x + 12 = 0

x(x – 3) – 4(x – 3) = 0

(x – 4)(x – 3) = 0

x = 4, 3

y^{2} – 3y + 2 = 0

y^{2} – 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**

**y ^{2}= 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

y^{2} = 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.**x**^{2}+ 9x + 20 = 0

** y ^{2}+ 13y + 42 = 0**

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

**Ans.**

x^{2} + 9x + 20 = 0

x^{2} + 4x + 5x + 20 = 0

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

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

x = -5, -4

y^{2} + 13y + 42 = 0

y^{2} + 6y + 7y + 42 = 0

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

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

y = -7, -6

x > y

**x**^{2}– 18x + 72 = 0

** y ^{2}– 21y + 108 = 0**

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

**Ans.**

x^{2} – 18x + 72 = 0

x^{2} – 12x – 6x + 72 = 0

x(x – 12) – 6(x – 12) = 0

x = 6, 12

y^{2} – 21y + 108 = 0

y^{2} – 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.

**2x**^{2}– 2x – 84 = 0

**y ^{2}– 2y – 48 = 0**

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

**Ans.**

2x^{2} – 2x – 84 = 0

2x^{2} – 14x + 12x – 84 = 0

2x(x – 7) + 12(x – 7) = 0

(2x + 12)(x – 7) = 0

x = -6, 7

y^{2} – 2y – 48 = 0

y^{2} – 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

**x**^{2}+ 32x – 144 = 0

** y ^{2}– 31y + 108 = 0**

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

**Ans.**

x^{2} + 32x – 144 = 0

x^{2} + 36x – 4x – 144 = 0

x(x + 36) – 4(x + 36) = 0

(x – 4)(x + 36) = 0

x = 4, -36

y^{2} – 31y + 108 = 0

y^{2} – 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.**x**^{2}+ 2x – 195 = 0

** y ^{2}+ y – 182 = 0**

Ans. x^{2} + 2x – 195 = 0

x^{2} + 15x – 13x – 195 = 0

x(x + 15) – 13(x + 15) = 0

(x – 13)(x + 15) = 0

X = 13, -15

y^{2} + y – 182 = 0

y^{2} + 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.**x**^{2}– 25x + 136 = 0

** y ^{2}– y – 56 = 0**

Ans. x^{2} – 25x + 136 = 0

x^{2} – 17x – 8x + 136 = 0

x(x – 17) – 8(x – 17) = 0

(x – 8)(x – 17) = 0

x = 8, 17

y^{2} – y – 56 = 0

y^{2} – 8y + 7y – 56 = 0

y(y – 8) + 7(y – 8) = 0

(y + 7)(y – 8) = 0

y = -7, 8

Hence, x ≥ y

3.**2x**^{2}+ 16x + 30 = 0

** y ^{2}+ 6y + 8 = 0**

Ans. 2x^{2} + 16x + 30 = 0

2x^{2} + 10x + 6x + 30 = 0

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

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

x = -3, -5

y^{2} + 6y + 8 = 0

y^{2} + 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. x^{2} + 5x + 6 = 0

x^{2} + 3x + 2x + 6 = 0

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

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

x = -2, -3

y^{2} – 2y + 1 = 0

y^{2} – y – y + 1 = 0

y(y – 1) – 1(y – 1) = 0

(y – 1)(y – 1) = 0

y = 1, 1

Hence, x < Y

**x**^{2}– 11x + 30 = 0

**y ^{2}– 21y + 90 = 0**

Ans. x^{2} – 11x + 30 = 0

x^{2} – 6x – 5x + 30 = 0

x(x – 6) – 5(x – 6) = 0

(x – 5)(x – 6) = 0

x = 5, 6

y^{2} – 21y + 90 = 0

y^{2} – 15y – 6y + 90 = 0

y(y – 15) – 6(y – 15) = 0

(y – 6)(y – 15) = 0

y = 6, 15

Hence, x ≤ y

## Methods to solve Quadratic Equation

There are a lot of methods in the books as well as recommended by youtube channels to solve the quadratic equation in less time. However, not every way of solving the quadratic equation problems are worth the time for an aspirant. In fact, since we will be studying for the competitive exam therefore it is very important that the candidates get hands-on methods to solve the roots of the quadratic equation that provides the answer in less time. Hence, with this idea, we are providing two major methods that have been used and followed by candidates who have previously aced the topic and scored good marks in the quantitative aptitude section.

**Sum of Roots of a Quadratic Equation**

This is the first and most widely used method to solve quadratic equation problems owing to its simple way and speedy calculations. A lot of students, however, get confused with the factors though therefore to help the students we have tried to explain the same with the help of an example as mentioned below:

x2 – 10x – 21 = 0 ——- (i)

x2 – 7x – 3x – 21 = 0

x(x – 7) + 3(x – 7) = 0

(x + 3)(x – 7) = 0

x = -3, 7

Therefore, -3 and 7 are the two roots of the above question.

**Quadratic Equation Using Quadratic Formula**

This is a complex and more complicated formula as compared to the previous one and includes following a quadratic formula as mentioned below.

For example, dx²+ex+c=0 is a quadratic equation, then D or Discriminant = e²-4dc

Now, if we apply the formula on any of the quadratic questions then you will get your answer,

x2 – 10x – 21 = 0

D = 10×10 – 4x1x21

D = 100-84

D = 16

The candidates can check their answer using the interpretation as given in the table below:

D>0 | Two roots of the Quadratic Equation are real and unequal |

D=0 | Two roots of the Quadratic Equation are real and equal. |

D<0 | No real Roots of the Quadratic Equation |

D>0 and D is a perfect square | The roots of the Quadratic Equation are Rational |

D>0 and D is not a perfect square | The roots of the Quadratic Equation are irrational |

## Books to ace the Quadratic Problems

Quadratic Equation is one of that topic of the Quantitative Aptitude section that can easily help a candidate in fetching 4 to 5 marks easily. However, this is only possible if the candidates are well versed with the correct technique required to solve the quadratic equation problems. One of the best ways to have a hold on the same is following books that are well versed with the demands of the exam. Hence, to help the candidates with the concept as well as sample questions for practice. We have compiled a list of books that have been helping a lot of candidates with the same:

- QUADRATIC EQUATIONS: RESOLVED
- Vedic Mathematics Sutra
- Mathematical Formulae by N.K. Singh
- Guidely Topic set on Quadratic Equation

## Reasons to prepare Quadratic Equation with Guidely

There are a lot of reasons citing which the candidates should only prepare for the quadratic equation using the resources of Guidely. Losing out on quadratic equation questions is not a very good idea for the aspirants. The most important and followed reasons are listed below for helping the candidates:

- The quadratic equations of different levels i.e. easy, moderate and high are solved and presented to the candidates.
- The candidates can also follow the various quadratic equation tricks to give an extra speed for their calculation while practising at home as well as during exam time.
- The practice PDF of quadratic equation problems are developed by keeping in mind the level of the questions asked in the different exams i.e. banking/ssc etc.
- The candidates can also follow the quizzes and tests developed by subject matter expert to prepare well for the quantitative aptitude section and especially the quadratic equation problems.
- The candidates can take the help of the youtube channel of Guidely in which the expert faculties conduct practise sessions as well as doubt classes to cover single doubts of the candidates regarding the roots of the quadratic equation problems.

## Practice More Quadratic Equation

More 1000+ Quadratic Equation Practice Questions