Basics of Quadratic Equations for Bank Exams

Quadratic Equation is one of the most favourite topics of almost every banking exam. A set of 4-5 Questions on Quadratic Equations are asked in Quantitative Aptitude section of many Banking and Insurance Examinations. Generally, two quadratic equations in two different variables are given.

We have to solve both of the Quadratic equations to get to know the relation between both the variables.

Suppose we have two variables ‘x’ and ‘y’. The relationship between the variables can be any one of the following:

a)   x>y

b)   x

c)    x=y or relation can’t be established between x & y

d)    x≥y

e)    x≤y

Methods of finding roots of a quadratic equation

First method:

The general quadratic equation is

ax²+ bx + c = 0

or, x²+(b/a)x+(c/a)=0

Now let us compare these two highlighted equations,

After comparison, we will get:

(α+β) = -(b/a)

αβ = c/a

Example:

x²+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 solution for equation x²+9x+20=0.

Second method:

x²+(4+5)x+(4*5)=0

x²+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:

Use of formula for finding the roots of a quadratic equation:

x=[-b± √{b²-4ac}]/2a

x=[-9± √{9²-4*1*20}]/2*1

x=[-9± √{81-80}]/2

x=[-9± √1]/2

x=[-9± 1]/2

x=(-9+1)/2 and x=(-9-1)/2

x=-8/2 and x=-10/2

x=-4 and x=-5

Any of these three methods can be used to find out the roots of a quadratic equation.

Questions asked on quadratic equations in exam

Direction: In the following questions, there are two equations (I) and (II). Solve the equations and answer accordingly:

Q.1) I. x² - 5x + 6 = 0

II. y²+y - 6 = 0

Solution:

Solving equation I:

x² - 5x + 6 = 0

x² - 3x - 2x + (3*2) = 0

x(x-3) -2(x-3) = 0

(x-3)(x-2) = 0

x =2 and 3

Solving equation II:

y² +y - 6 = 0

y² +3y-2y-6 = 0

y (y+3) -2(y+3) = 0

(y+3)(y-2)=0

y=-3 and 2

We can clearly see using number line that x is greater than y but x and y have a common point. So the relationship between x and y will be x≥y.

Q.2) I. x² +2x -3 = 0

II. y²+7y + 12 = 0

Solution:

Solving equation I:

x² +2x-3 = 0

x² +3x – x -3 = 0

x(x+3) -1(x+3) = 0

(x+3)(x-1) = 0

x =-3 and 1

Solving equation II:

y² +7y + 12 = 0

y²+4y+3y + 12 = 0

y(y+4)+3(y+4)=0

(y+3)(y+4)=0

y=-4 and -3

We can clearly conclude from the number line that x≥y is the relation between x and y.

Q.3) I. x² = 49

II. y = √49

Solution:

Solving equation I:

x² = 49

x =-7 and 7

Solving equation II:

y = √49

y=7

So we can clearly conclude using number line that y≥x.

Q.4) I. x² +7x +10 = 0

II. 2y²-7y + 6 = 0

Solution:

Solving equation I:

x² +7x+10 = 0

x² +5x+2x+10 = 0

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

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

x =-5 and -2

Solving equation II:

2y² -7y + 6 = 0

2y² -4y-3y + 6 = 0

2y(y-2)-3(y-2)=0

(y-2)(2y-3)=0

Y=3/2 and 2

y=1.5 and 2

So we can clearly conclude using number line that y>x.

Q.5)
I.8x² – 23x + 15 = 0
II. 3y² + 11y + 8 = 0
A. if x > y 
B. if x ≤ y 
C. if x ≥ y 
D. if x < y
E. if x = y or relationship between x and y can't be established

Solution:

Solving equation I:

I. 8x² – 23x + 15 = 0
8x² – 8x – 15x + 15 = 0
8x (x – 1) – 15 (x – 1) = 0
(8x – 15) (x – 1) = 0
x = 15/8 , 1

II. 3y² + 11y + 8 = 0
3y² + 3y + 8y + 8 = 0
3y (y + 1) + 8 (y + 1) = 0
(3y + 8) (y + 1) = 0
 y = -8/3 , -1 
Hence, x > y
Hence, option A is correct.

Key points related to Quadratic Equations:

1. Use one of the three methods to find out the roots of both the equations one by one.

2. After finding out roots, draw them on the number line.

3. On the number line, 5 possibilities can be there:

• x ends well before starting of y then the relation will be x

• y ends well before starting of x then the relation will be y

• y starts exactly at the same point where x ends then the relationship will be x≤y.

• x starts exactly at the same point where y ends then the relationship will be x≥y.

• y starts before ending of x or vice-versa then no relation can be established between x & y