# Quadratic Equations - Aptitude Tricks

In this article, we provide various techniques to solve aptitude questions on quadratic equations with the help of tips and tricks. A quadratic equation has at most two solutions. Mathematically the solutions of a quadratic equation are called the 'roots' of the equation. There are possibilities to get real or complex roots for a quadratic equation.

In this topic we are considering the following concepts.
• Sum and product of the roots.
• Nature of roots.
• Construction of a quadratic equation.
• Maximum and minimum value of a quadratic expression.

An equation of the form ax2 + bx + c = 0, where a, b and c are real values and a ≠ 0, is called a quadratic equation in variable 'x'.
More generally if P(x) is a second degree polynomial, then P(x) = 0 is a quadratic equation.
Ex: x2 + 2x + 3 = 0
2x2 - x + 1 = 0

Solve a quadratic equation means to find the values of the variable.

### Method I: Compare with corresponding algebraic identities.

 Recall the following algebraic identities: k (a + b) = ka + kb (a+b) ( a - b) = a2 - b2 (a + b)2 = a2 + 2ab + b2 (a - b ) 2 = a2 - 2 ab + b2 (x + a) (x + b) = x2 + (a + b)x + ab

Examples:

• x2 - x =0
Solution:
x2 - x =0
x(x-1) =0
x = 0 or x-1 =0
x = 0 or x =1
The roots are 0 or 1

• 64a2 - 9 =0
Solution: (8a)2-(3)2 ≥ 0
(8a+3)(8a-3) = 0
8a+3 = 0 or 8a-3 = 0
a=-3/8 or a = 3/8
Roots are -3/8 , 3/8

• x2 + 8x + 16 = 0
Solution:
(x+4)2 ≥ 0
x + 4 =0
x = - 4
Root = - 4

• y2 - 6y + 9 = 0
Solution:
(y - 3)2 = 0
y - 3 = 0
y = 3
Root = 3

• b2 + 5b + 6 = 0
Solution:
(b+2)(b+3) = 0
b+2 = 0 or b+3 = 0
b = - 2 or b = -3
Roots = - 2, - 3

• b2 - b - 6 = 0
Solution:
(b - 3)(b+2) = 0
b - 3 = 0 or b+2 =0
b = 3 or b = - 2
Roots = - 2, 3

• x2 - 8x + 15 = 0
Solution:
(x-3)(x-5) = 0
x - 3 = 0 or x - 5 =0
x = 3 or x = 5
Roots = 3, 5

• x2 - 8x + 20 = 0
Solution:
x2 - 8x + 16 + 4 = 0
(x - 4)2 + 4 = 0
(x - 4)2 = - 4
x - 4 = √(- 4)
x = 4 + √(- 4)
This equation doesn't have a real root, because a negative values under the root is an imaginary number
i.e. = 4i
Root = 4 + 4i
= 4(1 + i)

### Method 2: Splitting the middle term

In a quadratic equation of the form ax2 + bx + c = 0, if a≠1 then we are normally applying the method 'splitting middle term'.

Examples:

• 2x2 + 11x + 9 = 0

Solution:

How to split the middle term 11x ?
Consider two quantities p and q. Just compare the coefficient of x from the given equation as the sum of p and q.
i.e. p + q = 11
Compare the product of the coefficient of x2 and the constant term as the product of p and q.
i.e. p * q = 2* 9 = 18
Find such two values p and q such that p + q = 11 and pq = 18
p = 2 , q = 9
i.e. 11x can split as 2x + 9x
2x2 + 11x +9 = 0
2x(x+1) + 9(x+1) = 0
(2x+9)(x+1) = 0
2x + 9 = 0 and x + 1 = 0
x = -9/2 and x=-1
Roots = -9/2 , -1
• 3x2+17x+10 = 0
3x2 + 15x + 2x + 10 = 0
3x(x+5)+2(x+5) = 0
(3x+2)(x+5) = 0
3x + 2 = 0 and x + 5 = 0
x= -2/3 and x=-5
Roots = -2/3 , -5

Easy Method:

We already found the suitable value for p and q. Then the roots of the equation can find easily from the value of p and q.

Roots are -p/a and -q/a , where a is the coefficient of x2

In the above example p =15 and q = 2 so the roots are -15/3 = -5 and -2/3

• 2x2 - 3x-9=0

Solution:

pq = 2 * (-9) = -18
Product of two vales is negative, therefore p and q from opposite signs.
p - q = -3
so, p = -6
q = 3
Roots are -p/2 and -q/2
-p/2 = -(-6)/2 = 3
-q/2 = -(3)/2 = -3/2

Or

2x2 - 6x+3x-9=0
2x(x-3)+3(x-3) = 0
(2x+3)(x-3) = 0
2x+3=0 or x-3 =0
x=-3/2 or x=3
Roots = -3/2 , 3

Standard form of a quadratic equation is ax2 + bx + c = 0

x = ( - b ± √(b2 - 4ac) ) / 2a

Examples:

• x2+7x+3 = 0

Solution:

a=1
b=7
c=3
x = ( - b ± √(b2 - 4ac) ) / 2a
x = ( - 7 ± √(72 - 4 * 1 * 3) ) / 2 * 1
x = ( - 7 ± √(49 - 12) ) / 2
x = ( - 7 ± √ 37 ) / 2
Roots are x = ( - 7 + √ 37 ) / 2 and x = ( - 7 - √ 37 ) / 2

## Sum and product of the roots

 If ax2 + bx + c = 0 is a quadratic equation in x. Then, Sum of the roots = -b/a Product of the roots = c/a Therefore the standard form of a quadratic equation can consider as; x2 - (sum of the roots) x + product of the roots = 0

E.g. x2 - 5x + 6 = 0

Compare it with the standard form

x2 - (sum of the roots) x + product of the roots = 0

i.e. sum of the roots = 5

Product of the roots = 6

Hence the roots are 2 and 3

## Nature of roots

A quadratic equation has at most two real roots. It means some quadratic equation has one real root (two identical real roots) or some has no real roots, then they have the roots in the form of complex numbers. Whenever the equation has one or two real roots again it can be rational or irrational. These kinds of all different natures of the roots of a quadratic equation can find without solving the equation.

The nature of the roots of a quadratic equation is fully depends on the value of the expression 'b2 - 4ac'. This called the discriminant of the equation.

As per the different nature values of the discriminant the nature of the roots can be classified in the following manner.

Condition I:

If b2 - 4ac > 0 then there are two situations can be considered.

i. If b2 - 4ac is a perfect square then the equation has two distinct rational roots.

E.g. x2-2x-8 =0

a=1 b=-2 c=-8

x = ( - b ± √(b2 - 4ac) ) / 2a

= ( - (-2) ± √((-2)2 - 4 * 1 * 8) ) / 2 * 1

x = ( 2 ± √(4 + 32) ) / 2

x = ( 2 ± 6) / 2

(2 + 6)/2 or (2 -6)/2

4 or -2

Here b2 - 4ac = 36, it is greater than zero and a perfect square too.

The equation has two distinct rational roots.

ii. If b2 - 4ac is not a perfect square then the equation has two distinct irrational roots. If one of the roots in the form m + √n , then the other root in the form m-√n, where m and n are two rational numbers and n is not a perfect square.

E.g. x2 + 4x + 2 = 0

a = 1 b = 4 c = 2

x = -4 ± √(16 -4 * 1 * 2) / 2

= -4 ± √8 / 2

= -4 ± 2√2 / 2

= -2 ± 2√2

The roots are -2 + 2√2 and -2 - 2√2

Here b2 - 4ac = 8, it is greater than zero and it is not a perfect square. Hence the equation has two distinct irrational roots.

Condition II:

If b2 - 4ac = 0, then the equation has two identical (equal) rational roots.

E.g. x2+6x +9 = 0

a = 1 b = 6 c = 9

x = -6 ± √(36 - 4 * 1 * 9) / 2

= -6 ± √0 / 2

= -6 / 2

= -3

The roots are -3 and -3

i.e. There is only one rational root or the roots are equal.

Here b2 - 4ac = 0. Hence the equation has two equal rational roots.

Condition III:

If b2 - 4ac < 0, then the equation has two distinct complex roots in the form m+ni and m-ni

E.g. x2+2x +3 = 0

a = 1 b = 2 c = 3

x = -2 ± √(4 - 4 * 1 * 3) / 2

= -2 ± √8 / 2 = (- 2 ± 2√2 * i) / 2

= -1 ± √2 * i

The roots are -1 + √2 * i and -1 - √2 * i

## Construction of quadratic equation as per the condition on roots

• Roots are given

If the roots of the quadratic equation are given, then it is possible to frame the quadratic equation.

i.e. If the two distinct roots of a quadratic equation are and b, then the quadratic equation is,

(x-a)(x-b) = 0

x2 - (a + b) + ab = 0

Or

Use the general form x2 - (sum of roots) x + product of roots = 0

E.g. If 2 and -3 are the roots of a quadratic equation then find the equation?

Solution:

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

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

x2+3x-2x-6 = 0

x2 + x - 6 = 0

or

Sum of roots = 2-3 = -1

Product of roots = 2*(-3) = -6

Then the equation is x2-(-1) x+ (-6) = 0

i.e. x2 + x - 6 = 0

E.g. If one of the two distinct roots of a quadratic equation is 4+√5 then find the equation?

Solution:

If one of the two roots is 4+√5 then the other root is 4-√5

Sum of roots = (4+√5) + ( 4-√5)

= 8

Product of the roots = (4+√5) *( 4-√5)

= 16 - 5

= 11

The equation is

x2 - 8x + 11 = 0

E.g. If a quadratic equation has two identical roots and each roots equal to -7 then find the equation?

Solution:

Both the roots are equal to -7

Equation is (x-(-7)) 2 = 0

(x + 7)2 = 0

x2 + 14x + 49 = 0

• Relation between the roots are given

If there is sufficient relation between the roots of an equation is given then the equation can find easily.

Examples:

1. Sum of the roots of an equation is -5 and their product is 6, what is the quadratic equation?

Solution:

Sum of roots = -5

Product of roots = 6

Equation is x2 - (sum of roots)x+ product of roots = 0

i.e. x2 - (-5) x + 6 = 0

x2 +5x + 6 = 0

2. If the sum of the roots is 19 and the difference of the roots is 5 then find the quadratic equation?

Solution:

Let the roots are a and b.

a + b = 19

a - b = 5

Solving equations, then a = 12 and b = 7

The equation is (x-12) (x-7) = 0

x2 -19x + 84 = 0

3. If the product and difference of two distinct roots of a quadratic equation are 15 and 2 respectively, then what is the equation?

Solution:

Let the roots are a and b.

a - b = 2

ab = 15

(a-b)2 + 4ab = (a + b)2

22 + 4 * 15 = (a+b)2

(a+b)2 = 64

a+b = ± 8

Then there are two equations are possible.

x2 -8x + 15 = 0 and x2 +8x + 15 = 0

4. If the ratio between the roots of a quadratic equation is 1:4 and their sum is 10. Find the equation?

Solution:

a:b = 1:4

a=1k and b = 4k

a + b = 1k + 4k = 5k

5k = 10

k = 2

a = 2 and b = 8

ab = 2 * 8 = 16

Hence the equation is x2 -10x + 16 = 0

• Roots are reciprocals of the root of another equation

If the root of a quadratic equation are the reciprocals of the root of another quadratic equation of the form ax2 + bx + c = 0 the equation will get through the interchanging of the coefficient of x2 and constant term in ax2+bx+c = 0.

So , the required equation is cx2 + bx + a = 0

If the roots of ax2 + bx + c = 0 are p and q, then the roots of the equation cx2 + bx + a = 0 are 1/p and 1/q

E.g. find the quadratic equation whose roots are reciprocals of the roots of 2x2 + 9x + 7 = 0

Solution:

2x2 + 9x + 7 = 0

2x2 + 2x +7x + 7 = 0

2x(x + 1) + 7(x + 1) = 0

(2x + 7)(x + 1) = 0

x = -7/2 x = -1

Roots of the given equations are -7/2 and -1

Roots of the required equations are -2/7 and -1

Hence the equation is

(x - 2/7) * (x - 1) = 0

(x + 2/7) * (x + 1) = 0

x2 + 9/7 x + 2/7 = 0

Multiply the LHS and RHS by 7

Then 7x2 + 9x + 7 = 0

• Roots are 'k' more than the roots of another equation

The roots of a quadratic equation in x, ax2 + bx + c = 0 are p and q , then a quadratic equation whose roots are p + k and q + k can be found in the following way.

Consider the required quadratic equation in y.

Then x + k = y

i.e. x = y - k

For framing the required equation just substitute 'y-k' in place of 'x' of the given equation

e.g. Find the equation whose roots are 2 more than the roots of x2 - 3x - 28 = 0

Solution:

Let the roots of the given equation be 'x'. Then the roots of the required equation (in y) is

y = x+2

x = y - 2

Substitute 'y-2' in place of 'x' of the given equation.

i.e. (y-2)2 - 3(y-2) - 28 = 0

y2 - 4y + 4 - 3y + 6 - 28 = 0

x2 - 7y - 18 = 0

Verification

Roots of x2 - 3x - 28 = 0 are 7 and -4

Roots of y2 - 7 y - 18 = 0 are 9 and -2

7 + 2 = 9 and - 4 + 2 = -2

Hence it is verified.

i.e. roots of y2 - 7y - 18 = 0 are 2 more than the roots of x2 - 3x - 28 = 0.

• Roots are k less than the roots of a quadratic equation

If the roots of a quadratic equation ax2 + bx + c = 0 in x, then the roots of required equation in y is

y = x - 2

x = y + 2

For getting this equation substitute y + 2 in place of x in the given equation

• Roots are k times the roots of another equation

If the roots of a quadratic equation ax2 + bx + c = 0 are p and q, then the quadratic equation whose roots are 'kp' and 'kq' can be found in the following way.

Let the roots of equation ax2 + bx + c = 0 in x, then the roots of the required equation (in y ) is

y = kx

x = y/k

For getting the equation, substitute y/k in place of 'x' in the given equation.

Example:

Find the quadratic equation whose roots are 3 times the roots of x2 - 5x + 6 = 0

Solution:

In the new equation x = y/3

The equation is (y/3)2 - 5(y/3)+6 = 0

y2/9 - 5y/3 + 6 =0

Multiply the equation by 9

y2 -15y + 54 = 0

Verification:

Roots of x2-5x+6 = 0 are 2 and 3

Roots of y2-15y+54 = 0 are 6 and 9

2*3 =6 and 3*3 = 9

Hence it is verified.

i.e. roots of y2-15y+54 = 0 3 times the roots of x2-5x+6 = 0.

## Maximum or minimum value of a quadratic expression

The quadratic expression in x , ax2+bx+c will take different values as per the different real values for the variable x.

• The quadratic expression ax2+bx+c has a minimum value when a>0 at x=-b/2a and this minimum value is (4ac - b2) / 4a
• The quadratic expression ax2+bx+c has a maximum value when a<0 at x=-b/2a and this maximum value is (4ac - b2) / 4a

Note:

Refer the maximum and minimum value of a quadratic function, from the topic 'Functions and Graphs'