Hello

Quadratic Equation is one of the most favorite topics of almost every banking exam. Questions on Quadratic Equations are asked in Quantitative Aptitude section. 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:

- x>y
- x<y
- x=y or relation can’t be established between x & y
- x≥y
- x≤y

__Meaning of different symbols__

__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 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 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 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 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 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 Equation__

__General form of a Quadratic Equation__

**ax ^{2} + bx + c = 0**

Quadratic equation means that it will definitely have the maximum power of the variable as ‘2’ which means **we will always see ax ^{2} term **in a quadratic equation.

Or we can say that b can be 0, c can be 0 but **a will never be 0.**

Whenever we solve a quadratic equation, we will get exactly 2 values of the equation. These 2 values are called **roots **of the equation. The roots of the equation always satisfy the equation. So in case of doubt, we can check the solution by putting the values back into the equation. If the equation turns out to be zero then our roots are correct.

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 (x-α)=0 and (x-β)=0

If we multiply both the equations, we will get

(x-α)*(x-β)=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__

__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 us compare these two highlighted equations,

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.

Answer is -5 and -4. So these two are the roots or solution 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:__

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

**x=[-b± √{b ^{2}-4ac}]/2a**

x=[-9± √{9^{2}-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.

__Example-1:__ 2**x ^{2 }+11x + 9 = 0**

__Solution:__

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

2x(x+1) +9(x+1) = 0

(x+1)(2x+9)=0

x = -9/2 = -4.5 and x = -1

__Example-2:__ 4**x ^{2 }+7x -2 = 0**

__Solution:__

**4****x ^{2 }+7x -2**

x=[-b± √{b^{2}-4ac}]/2a

x=[-7± √{7^{2}-4*4*-2}]/2*4

x=[-7± √{49+32}]/8

x=[-7± √81]/8

x=[-7± 9]/8

x=(-7-9)/8 and x=(–7+9)/8

x=-16/2 and x=2/8

x = -2 and ¼

x=-2 and 0.25

__Actual questions asked on quadratic equations in exam__

__Actual 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. x ^{2 }– 5x + 6 = 0**

**II. y ^{2}+y – 6 = 0**

**Solution:**

**Solving equation I:**

x^{2 }– 5x + 6 = 0

x^{2 }– 3x – 2x + (3*2) = 0

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

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

x =2 and 3

**Solving equation II:**

y^{2} +y – 6 = 0

y^{2} +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. x ^{2 }+2x -3 = 0**

**II. y ^{2}+7y + 12 = 0**

**Solution:**

**Solving equation I:**

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

x^{2 }+3x – x -3 = 0

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

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

x =-3 and 1

**Solving equation II:**

y^{2} +7y + 12 = 0

y^{2}+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. x ^{2 }= 49**

**II. y = √49**

**Solution:**

**Solving equation I:**

x^{2 }= 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. x ^{2 }+7x +10 = 0**

**II. 2y ^{2}-7y + 6 = 0**

**Solution:**

**Solving equation I:**

x^{2 }+7x+10 = 0

x^{2 }+5x+2x+10 = 0

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

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

x =-5 and -2

**Solving equation II:**

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

2y^{2} -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. x ^{2 }+7x +10 = 0**

**II. 2y ^{2} = 50**

**Solution:**

**Solving equation I:**

x^{2 }+7x+10 = 0

x^{2 }+5x+2x+10 = 0

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

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

x =-5 and -2

**Solving equation II:**

2y^{2} = 50

y^{2} = 25

y=-5 and 5

As we can see in the number line that x and y values have a common area so no relation can be established between x & y.

**Key points related to Quadratic Equations:**

- Use one of the three methods to find out the roots of both the equations one by one.
- After finding out roots, draw them on the number line.
- On the number line, 5 possibilities can be there:

- x ends well before starting of y then the relation will be x<y.
- y ends well before starting of x then the relation will be y<x.
- 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.

Thanks

## Leave a Reply