# Quadratic Equation PDF For All Competitive Exam

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Quadratic Equation PDF: Quadratic equation is one such topic that is asked in almost all the banking exams. Approximately five questions are asked from this topic which carries 5 marks and you can easily fetch those five marks if you solve this topic with the right approach. In this article, we will be telling you the correct way of solving the questions from this topic and if you want to make sure that you master this topic then readout this article.

Also Check: Score Up PDF For the Quantitative Aptitude Section

## Quadratic Equation PDF For All Competitive Exams

Quadratic Equation Questions are very important for every Banking, Insurance, SSC, Railway, and other government exams. This topic requires a lot of practice and the right approach to scoring good marks in this section. To help you with your practice we have brought you a free PDF on Quadratic Equation Questions with detailed solutions For practice. Download the free Quadratic Equation PDF.

In quadratic equation, students are expected to find the relation between two variable given in the two equation. The related can be of Smaller to, Greater to, Greater than, Smaller than and Equal to.

Linear Equation– In this type, when a student solve the equation then he/she will only get one value for X and for Y.

for ex- 2X+3Y=5, X+2Y=6

2X+3Y=5—–(eq.1), (X+2Y=6)X2——(eq.2)

Solving these two equations, we will get

We get, X=-8 & Y= 7

Hence Y>X, this will be our final answer

Squares Equation– In this type of question, we have to find the sqaure root of the given below and we end up getting two values each for X and Y but one is negative while other is positive. The best thing about such type of questions is that the answer to such questions are always Cannot be Determined.

For e.q., X= 1600 & Y=2500

On finding their square root we will get X=+40, -40 & Y= +50, -50

Hence the answer will be Cannot be determined.

Sqaures and Sqaure root equation– In this type of question, one is sqaure while the other one is square root and we know that sqaure root always gives positive value when we solve it.

For e.g. X=1600 & Y= 2500

On solving them we will get, X= +40, -40 & Y=+50

Hence, our answer will be Y>X

Cube Cases– In this type of questions, the cube will be given and you have to answer the relation between those cubes.

For e.g. X= 1331 & Y= 729

When we solve these two equation, we will get X= 11 & Y= 9

Hence, X>Y. The trick to solve these questions is that whichsoever cube will be greater when you will solve it that cube will remain greater.

### Table Method to Solve Quadratic Equation

This is one of the best methods to solve Quadratic Equation questions. You can answer the question without actually solving the equations.

For e.g. x2 – 7x + 10 = 0, y2 + 8y + 15 = 0

We can solve this equation without actually solving the equation. Let’s see

• First look at the sign in equation.1 which are -, + means its roots will be +,+ as per our table.
• Secondly, look at the other equation, we will see the signs are +, + which means that our roots will be -, -.
• Hence X>Y as it has both the roots positive.

### Different cases in Quadratic Equations

Directions (01-05): In the following questions, two equations I and II are given. You have to solve both equations and give answer as,

1) I) 3x2 + 30x + 63 = 0

II) y2– 4y – 21 = 0

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x=y or the relation cannot be established.

2) I) x2 – 33x + 162=0

II) y2– 30y + 144=0

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x=y or the relation cannot be established.

3) I) x2 + 23x + 130 = 0

II) 2y2+ 32y + 120 = 0

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x=y or the relation cannot be established.

4) I) x2 – 17x + 52=0

II) y2– 7y – 60=0

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

5) I) x2 + 15x – 184=0

II) y2– 30y + 176=0

a) x > y

b) x ≥ y

c) x < y

d) x ≤ y

e) x = y or the relation cannot be established.

3x2 + 30x + 63 = 0

3x2 + 21x + 9x + 63 = 0

3x(x + 7) + 9(x + 7) = 0

(3x + 9)(x + 7) = 0

x = -3, -7

y2 – 4y – 21 = 0

y2 – 7y + 3y – 21 = 0

y(y – 7) + 3(y – 7) = 0

(y – 7)(y + 3) = 0

y = 7, -3

x ≤ y

x2 – 33x + 162 = 0

x2 – 27x – 6x + 162 = 0

x(x – 27) – 6(x – 27) = 0

(x – 6)(x – 27) = 0

x = 6, 27

y2 – 30y + 144=0

y2 – 24y – 6y + 144=0

y(y  – 24) – 6(y – 24)=0

(y – 6)(y – 24)=0

y = 6, 24

Relationship between x and y cannot be established.

x2 + 23x + 130=0

x2 + 13x + 10x + 130=0

x(x + 13) + 10(x + 13)=0

(x + 10)(x + 13)=0

x = -10, -13

2y2 + 32y + 120=0

2y2 + 20y + 12y + 120=0

2y(y + 10) + 12(y + 10)=0

(2y + 6)(y + 10)=0

y = -3, -10

x ≤ y

x2 – 17x + 52=0

x2 – 13x – 4x + 52=0

x(x – 13) – 4(x – 13)=0

(x – 4)(x – 13)=0

x = 4, 13

y2 – 7y – 60=0

y2 – 12y + 5y – 60=0

y(y – 12) + 5(y – 12)=0

(y + 5)(y – 12)=0

y = -5, 12

Relationship between x and y cannot be established.

x2 + 15x – 184=0

x2 + 23x – 8x – 184=0

x(x + 23) – 8(x + 23)=0

(x – 8)(x + 23)=0

x = 8, -23

y2 – 30y + 176=0

y2 – 22y – 8y + 176=0

y(y – 22) – 8(y – 22)=0

(y – 8)(y – 22)=0

y = 8, 22

x ≤ y