Basics of Quadratic Equations for Bank Exams 2022
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Almost every banking test includes a section on quadratic equations since it is one of the most important topics. Many Banking and Insurance Examinations include a series of 45 questions on quadratic equations in the Quantitative Aptitude exam. Two quadratic equations in two separate variables are provided in most cases.
Table of content
What is a Quadratic Equation?
Quadratics can be defined as a polynomial equation of a second degree, which implies that it comprises a minimum of one term that is squared. It is also called quadratic equations. The general form of the quadratic equation is ax² + bx + c = 0. where x is an unknown variable and a, b, and c are numerical coefficients.
To find the relationship between the two variables, we must solve both quadratic equations.
Let’s say we have two variables, x, and y. Any of the following relationships can exist between the variables:
 x>y
 x
 x=y or relation can’t be established between x & y
 x≥y
 x≤y
What is the Meaning of Different Symbols?
Before getting deep into the quadratic equations, let us try to understand the meaning of the basic operations used in finding the relationship between the variables –
(1) ‘>’ symbol: This symbol indicates that the variable on the left side is definitely greater than the variable on the right side of the symbol.
For example, x>y means x is definitely greater than y.
(2) ‘<’ symbol: This symbol indicates that the variable on the left is definitely smaller than the variable on the right side of the symbol.
For example x
(3) ‘=’ symbol: This symbol indicates that the variable on the left side is equal to the variable on the right side of the symbol.
For example, x=y means x is definitely equal to y.
(4) ‘≥’ symbol: This symbol indicates that the variable on the left side is either greater than or equal to the variable on the right side of the symbol.
For example, x≥y means x is either greater than y or equal to y.
(5) ‘≤’ symbol: This symbol indicates that the variable on the left side is either smaller than or equal to the variable on the right side of the symbol.
For example, x≤y means x is either smaller than y or equal to y.
General Form of a Quadratic Equationax^{2} + bx + c = 0 
The maximum power of the variable in a quadratic equation is always ‘2′, which means we will always get the ax2 term in a quadratic equation.
Or we can say that b can be 0, c can be 0 but a will never be 0.
We will always receive exactly two values of a quadratic equation when we solve it. The roots of the equation are these two values. The equation is always satisfied by the roots of the equation. In the event of doubt, we can doublecheck the solution by reentering the values into the equation. Our roots are right if the equation turns out to be zero.
Let us see how we can obtain a quadratic equation if we know the roots so that we will get a very clear concept of the basic formation of a quadratic equation.
Suppose we know both the roots as x=α and x=β.
Or we can say that (xα)=0 and (xβ)=0
If we multiply both the equations, we will get
(xα)*(xβ)=0
x^{2}– αx βx+ αβ=0
x^{2 }– (α+β)x+ αβ=0
The obtained equation is a quadratic equation having roots α and β.
Methods of Finding Roots of a Quadratic Equation
First method:
The general quadratic equation is
ax^{2}+ bx + c = 0
or, x^{2}+(b/a)x+(c/a)=0
Now let’s compare the two equations that have been indicated.
After comparison, we will get:
(α+β) = (b/a)
αβ = c/a
Example:
x^{2}+9x+20=0
a=1,b=9,c=20
(α+β) = 9/1 = 9
αβ = 20/1 = 20
So, now we have to think which two numbers multiplication gives us 20 and their addition gives 9.
The answer is 5 and 4. So these two are the roots or solutions for equation x^{2}+9x+20=0.
Second method:
x^{2}+(4+5)x+(4*5)=0
x^{2}+4x+5x+4*5=0
x(x+4)+5(x+4)=0
(x+4)(x+5)=0
So x=4 and x=5
Third method:
The following formula can be used to discover the roots of a quadratic equation:
x=[b± √{b^{2}4ac}]/2a
x=[9± √{9^{2}4*1*20}]/2*1
x=[9± √{8180}]/2
x=[9± √1]/2
x=[9± 1]/2
x=(9+1)/2 and x=(91)/2
x=8/2 and x=10/2
x=4 and x=5
Direction: In the following question two equations numbered I and II are given. You have to solve both equations and answer the question.
 I.X^{2} – 37X + 210 = 0
II. Y^{2} – 47Y + 280 = 0
A. X > Y
B. X ≥ Y
C. Y > X
D. Y ≥ X
E. X = Y OR the relationship cannot be established
Solution: The Answer is D.
X^{2} – 37X + 210 = 0
X^{2} – 30X –7X + 210 = 0
X(X – 30) – 7(X – 30) = 0
(X – 30)(X – 7) = 0
X = 7, 30
Y^{2} – 47Y + 280 = 0
Y^{2} – 40Y – 7Y + 280 = 0
Y(Y – 40) –7(Y – 40) = 0
(Y – 7)(Y – 40)=0
Y = 7, 40
X=30 is greater than Y=7 as well as less than Y=40
So, the relationship cannot be established
Direction: In the following question, there are two equations. Solve the equations and answer accordingly:
 I: x^{3 }= ()^{3}
II: 30y^{2 }= 750
A. x > y
B. x < y
C. x ≥ y
D. x ≤ y
E. x = y OR No relation can be established(CND)
Solution: The Answer is A.
x^{3}=()^{3}
x^{3}=(
x^{3}=216
x=
x=6
30y^{2}=750
y^{2}=750÷30
y^{2}=25
y=√25
y= ±5
x>y
Direction: In the following question two equations are given in variables X and Y. You have to solve these equations and determine the relation between X and Y.
 √(x + 20) = √256 – √121
y^{2 }+ 584 = 705
A. x > y
B. x < y
C. x ≥ y
D. x ≤ y
E. x = y OR No relation can be established(CND)
Solution: The Answer is E.
√(x + 20)=√256 – √121
√(x+20)=1611
√(x+20)=5
x+20=5^{2}
x+20=25
x=2520
x=5
y^{2}+584=705
y^{2}=705584
y^{2}=121
y = 11, 11
No relationship can be established between ‘x’ & ‘y’.
Directions: In each of the following questions two equations (I) and (II) are given. Solve both equations and give an answer.
 (I) 3x^{2}+ 8x + 4 = 0
(II) 4y^{2} – 19y + 12 = 0.
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or relationship cannot be established
Solution: The Answer is C.
(3x+2) (x+2)
Solving we get x= 2/3 or 2
(4y3) (y4)
y= 3/4 or 4
y>x
Direction: In the following question two equations numbered I and II are given. You have to solve both equations and mark the correct answer.
 I. X^{2 }– 11x + 28 = 0
II. y^{2 }+ y – 30 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. X = Y or the relationship cannot be established
Solution: The Answer is E.
I. X^{2}11x+28=0
X^{27x4x+28=0 }(x7)(x4)=0
x = 4, 7
II. y^{2}+y30=0
y^{2}+6y 5y30=0
(y+6)(y+5)=0
y = 5, 6
Relationships cannot be established
Any of these three methods can be used to find out the roots of a quadratic equation.
As we can see in the number line, x and y values have a common area, so no relation can be established between x & y.
You may Check other Quantitative Aptitude Study Notes
Here you will find study notes of all the chapters of the quantitative aptitude section for Bank Exams 2022 preparation. Kindly have a look at them.
S.No 
Topics 
1 

2  
3 
D.I Basics, Pie Chart, Table Chart, Bar Graph, Line Graph, Mixed Graph, Missing D.I 
4 

5 

6 

7 

8 

9 

10 

11 

12 

13 

14 

15 

16 

17 
Key Points Related to Quadratic Equations:
 Find the roots of both equations one at a time using one of the three approaches.
 Draw the roots on the number line after you’ve discovered them.
 There are 5 choices on the number line:
 If x comes to an end before y begins, the relationship will be x
 If y comes to an end before x, the relationship will be y
 If y begins at the same place where x finishes, the connection will be x≤y.
 If x begins and finishes precisely where y does, the relationship will be x≥y.
If y begins before x ends or vice versa, no relationship between x and y can be constructed.
Attempt daily Quizzes on Quantitative Aptitude. 
Quadratic equation questions do come in these bank exams too so it’s worth preparing this topic for sure.
1. SBI Clerk 2. SBI PO 3. IBPS Clerk 4. IBPS PO 5. IBPS RRB 7. RBI GradeB 9. Bihar State Cooperative Bank 10. SBI Apprentice 12. LIC AAO 13. FCI 
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