Chapter 6 ( Part 2) : Quadratic Equation

Definition: 
                        An equation whose degree is ‘2’ is called as quadratic equation.
                It can be written in the form of 
                ax² + bx + c = 0
                Here, a, b and c are real numbers 
                a ≠ 0     
Properties: 
1. A quadratic equation has two roots. (root can be called as the solution of quadratic equation)
2. These two roots individually when put in the quadratic equation it satisfies the given quadratic equation.
For e.g. for quadratic equation 2x² – 3x – 9,   x = 3 satisfies the given equation so therefore x = 3 is one of the root solution of the quadratic equation.
From the given equation nature of the roots can be found out such as real, imaginary, distinct etc.

A. Methods of finding roots of quadratic equation Solution of quadratic equation

1. By Factorization Method
2. By Formula Method
3. By Completing Square method

1. By Factorization Method

This method is based on the concept that if product of two numbers then at least one of the number is zero.
For quadratic equation,
ax² + bx + c = 0
we will expand the middle term ‘bx’ in two terms and thereafter taking common multiple out of it. The equation will be reduced to following step.
(px + q)(rx + s) = 0
Here, either (px + q) = 0 or (rx + s) = 0
So two roots (two value of x) can be found.
Let’s understand it by example.

Case–1   
when a = 1
x² + 11x + 21 = 0
we will find two multiplies of 21 in such a way that its addition or subtraction is 11.
x² + 11x + 21 = 0                                    
x² + 7x + 3x + 21 = 0                    
x(x + 7) + 3(x + 7) = 0               
(x + 7) (x + 3) = 0
x + 7 = 0   OR   x + 3 = 0
x = – 7     OR   x = – 3   

Case–2    
when a ≠ 1
3x² + 10x + 3 = 0 comparing with ax² + bx + c = 0,  a = 3, b = 10, c = 3.
3x² + 9x +1x + 3 = 0
3x (x + 3) + 1 (x + 3) = 0                               Here, a ≠ 1
(x + 3) (3x + 1) = 0                                       a × c = 3 × 3 = +9  [ +9 will be factored ]
x + 3 = 0   OR   3x + 1 = 0
x = -3     OR      x = (-1)/3                                                                                                         
  
2. By Formula Method

Solution of quadratic equation,
ax² + bx+ c = 0 can be find using the following formula,
x=(-b±√(b²-4ac))/2a
i.e. x=(-b+√(b²-4ac))/2a    OR  x=(-b-√(b²-4ac))/2a

Here, the term ‘b^2-4ac’ is called as ‘Discriminant’ and it defines the nature of root.
For e.g. x² - 15x + 54 = 0 
By comparing it with,  
ax² + bx + c = 0 
a = 1, b = - 15, c = 54
Let discriminant D = b²-4ac
                           = (-15)² – 4(1)(54)
                           = 225 – 216
                        D = 9
x =(-b±√(b²-4ac))/2a
   =(-15±√9)/(2(1))
   = (-15+√9)/2   OR   = (-15-√9)/(2(1))
   =(-6)/2            OR    = (-18)/2

x = -3                 OR  x = -9

3. By Completing Square Method

First method of solving quadratic equation will be factorization method but sometimes these are some equations in which we can not find the factors of constant term ‘c’ such that its sum is middle co-efficient ‘b’ such as,
x² + 9x + 2 = 0
x² + 21x + 11 = 0  etc.
To solve such problems we will use completing square method.
In these method,
We will add a suitable term to (ax² + bx) to make a completing square
For example x² + 8x + 2 = 0
In these equation if we add ‘16’ on the both the side of equation.

x² + 8x + 2 + 16 = 0 + 16
x² + 8x + 16 + 2 = 16
(x + 4)² + 2 – 16 = 0
(x + 4)² – 14 = 0
(x + 4)² – (√14 )² = 0
(x + 4 + √14 ) (x + 4 – √14) = 0
(x + 4 + √14) = 0   OR     (x + 4 – √14) = 0
 x = – 4 – √14        OR      x = – 4 + √14

How to find suitable term to make (ax² + bx) complete square.
Let ‘k’ be the term to add (ax² + bx) in order to make it complete square, then,
k =( b/2a )²
For e.g. x² + 6x + 7 = 0
Here a = 1, b = 6, c = 7
k =( b/2a )² =( 6/(2×1) )² = (3)² = 9
After adding and subtracting ‘9’ it will become
x² + 6x + 9 + 7 - 9 = 0
(x + 3)² + 7 – 9 = 0
(x + 3)² – 2 = 0
(x + 3)² – (√2)² = 0
(x + 3 + √2) ×  (x + 3 – √2) = 0
x = – 3 – √2    OR   x = – 3 + √2

B. Nature of roots and relations between them

Nature of roots for quadratic equations,
ax² + bx + c = 0
Depends upon the value of a, b and c
Discriminant = d = b² – 4ac
Case – 1     if d > 0 then the two roots will be real and distinct
Case – 2     if d = 0 then the two roots will be real and equal
Case – 3     if d < 0 then the two roots will be imaginary and distinct
For example what will be the nature of roots for quadratic equation,
x² + x – 6 = 0
Solution:
Comparing x² + x – 6 = 0 with ax² + bx + c = 0
a = 1, b = 1 and c = – 6
Discriminant = d = b² – 4ac
                         = 1 – 4(1)(–6)
                         = 1 + 24
                         = 25
As here d > 0, 
The given quadratic equation will have two real distinct roots.

SOME IMPORTANT CASES:

1.     When discriminant (d) = 0 then the roots of quadratic equation are, ∝ = β=(-b)/2a
2.     If a = c then roots will be reciprocal to each other
3.     If b = 0 then roots will be numerically equal but will have equal sign 
4.     If a and c are opposite sign then roots will have also opposite sign 
5.     For both the roots to be positive, (-b)/a > 0 and c/a > 0 and in the same way for both the roots to be negative b/a > 0 and c/a > 0.

C. Relation between roots and coefficients of Quadratic Equation

If ∝ and β are the roots of quadratic equation.
ax² + bx + c = 0
then,  
1. Sum of the roots = ∝ + β = (-b)/a
2. Product of the roots = ∝.β = c/a

D. Formation of Quadratic Equation:

Here we will form the quadratic equation when the roots (∝ and β) of a quadratic equation will be,
If ∝ and β are the two roots of a quadratic equation then,
(∝- β) (∝ + β) = 0   -> when two roots are given
x² - (∝ + β)x + ∝.β = 0    -> when relation between the roots are

For e.g. roots of quadratic are 2 and -3. The given equation will be?
Solution:
As we know when roots of quadratic equation are given the equation will be for ∝ = 2 and β = -3
(x - ∝) (x – β) = 0
(x - 2) (x + 3) = 0
x² + 3x – 2x – 6 = 0
x² + x – 6 = 0
This is the required quadratic equation.

E. Solution of different / other equation by Quadratic Equation:

There are many equation which are not quadratic equation and whose solution is much difficult but if we convert those equation by some substitution and simplification it can be easily solved. Some of them are solved below

Q.1. Solve the following  equation,
               x + 2/3x = 7/3

Solution:
Here the above equation is not quadratic equation but if we multiply the above equation by ‘x’ it becomes the quadratic equation,
x + 2/3x = 7/3
x (x + 2/3x) = 7x/3             …multiplying both side by ‘x’
x² + 2/3 = 7x/3
x² – 7x/3 + 2/3 = 0
3x² – 7x + 2 = 0                …multiplying both side by ‘3’
3x² – 6x – x  + 2 = 0                                                                     
3x(x – 2) – 1(x – 2) = 0
(3x – 1) (x – 2) = 0                                                                   
3x – 1 = 0    OR   x – 2 = 0
x = 1/3     OR   x = 2

Q.2. Solve the following equation 
Solution:

x⁴ – 13x² + 36 = 0
Here we can write, x⁴ = (x²)²
Let assumes, x² = m, so therefore equation becomes
m² – 13m + 36 = 0
m² – 9m – 4m + 36 = 0
m(m – 9) – 4( m – 9) = 0
(m – 9) (m – 4) = 0
m = 9     OR      m = 4
Now after resubstituting,
m = x²
x² = 9     OR      x² = 4
x = ± 3   OR       x = ± 2

Q.3. (x+3) (x+2) (x+4) (x+5) + 1 = 0

Solution:
The above equation is in the form of
(x+a) (x+b) (x+c) (x+d) + k = 0
To solve the above equation we have to make a pair of two terms
The pair will be selected on the following basis,
a + b = c + d    or
a + c = b + d    or
a + d = b + c
In our case,
(x+3) (x+2) (x+4) (x+5) + 1 = 0
{3 + 4 = 2 + 5}
(x+3) and (x+4) will make one pair and (x+5) and (x+2) will make another pair.
(x+3)(x+4)(x+2)(x+5) + 1 = 0
(x² + 4x + 3x + 12) (x² + 5x + 2x + 10) + 1 = 0
(x² + 7x + 12) (x² + 7x + 10) + 1 = 0
Let’s assume x² + 7x = t
(t+12) (t+10) + 1 = 0
t² + 10t + 12t + 120 + 1 = 0
t² + 22t + 121 = 0                                                                   
t² + 11t + 11t + 121 = 0
t(t+11) + 11(t+11) = 0
(t+11) (t+11) = 0                                                               
t + 11 = 0   OR    t + 11 = 0
t = - 11
After Resubstituting,
x² + 7x = - 11
x² + 7x – 11 = 0
By formula method,
a = 1, b = 7 and c = 11
b² – 4ac = (7)2 – 4(11)(1)
              = 49 – 44
              = 5
x=(-b±√(b²-4ac))/2a
  = (-7±√5)/2
x = (-7+√5)/2        OR    x = (-7-√5)/2

Q.4.  3(x² + 1/x² ) – 5 (x + 1/x) + 8 = 0
Solution:
As we know,
(x+1/x)² = x² + 2x. 1/x + 1/x² 
(x+1/x)² = x² + 2 + 2/x²
(x+1/x)² – 2 = x² + 2/x² 
Put (x+1/x) = m
3(m² – 2) – 5m + 8 = 0
3m² – 6 – 5m + 8 = 0
3m² – 5m + 2 = 0                                                                
3m² – 3m – 2m +2 = 0
3m(m – 1) – 2(m – 1) = 0 
m – 1 = 0      OR     3m – 2 = 0                                      
m = 1            OR     3m = 2
∴  m = 2/3 
After resubstituting
(x+1/x) = m
x+1/x = 1                and          x+1/x = 2/3
x² + 1 = x                                       x² + 1 = 2x/3                       …multiplying by ‘x’
x² – x + 1 = 0                                3x² + 3 = 2x
x² – x + 1 = 0                                3x² – 2x + 3 = 0
b² – 4ac:                                       b² – 4ac:
b² – 4ac=(–1)² –4(1)(1)                b² – 4ac = (–2)² –4(3)(3)
             = 1 – 4                                            = 4 – 36
             = –3                                                = – 32
As b² – 4ac < 0                        As b² – 4ac < 0
It has imaginary                       It has imaginary
Roots or no real                        Roots or no real roots.
Roots                                             

SOME IMPORTANT FORMULA

1.     a² + b² = (a+b)² – 2ab
2.     a³ + b³ = [(a+b)³ – 3ab (a+b)]
3.     (a – b)² = [(a + b)² – 4ab]
4.     a³ – b³ = [(a–b)³ + 3ab (a–b)]
5.     (a – 1/a)² = (a² + 1/a² ) + 2
6.     (a + 1/a)² = (a² – 1/a² ) + 2

F. Inequalities:

Quadratic inequality can be written in the following way,
ax² + bx + c > 0
ax²+ bx + c ≥ 0
ax² + bx + c < 0
ax² + bx + c ≤ 0

Steps to solve quadratic inequality:

What will be the solution of quadratic inequality
                               x² + 7x + 6 ≥ 0?
    Solution:
                x² + 7x + 6 = 0
                x² + 6x + 1x + 6 = 0  
                x² + 6x + 1x + 6 = 0
                x(x+6) +1 (x+6) = 0
                (x+1) (x+6) = 0
                x = –1  OR  x = –6
    
Now, we have to find interval on number line which satisfies the above conditions,
        
Let’s take three points 
          x = –7                           x = –4                              x = 1
    For x = –7                      For x = –4                          For x = 1
    x² + 7x + 6 ≥ 0            x² + 7x + 6 ≥ 0                   x² + 7x + 6 ≥ 0
   (–7)² +7(–7)+6 ≥ 0     (–4)² +7(–4)+6 ≥ 0              (1)² +7(1)+6 ≥ 0  
    49 – 49 + 6 ≥ 0           16 – 28 + 6 ≥ 0                    1+7+6≥ 0
             6  ≥ 0                       22 – 28 ≥ 0                    14 ≥ 06
                                                     -6 ≱ 0                      8  ≥ 0  
The solution is x > –1 and x < –6

Steps to solve Quadratic Inequality:

1.    Replace the inequality symbol with an equal sign
2.    Solve the quadratic equations
3.    Forms the number line and plot the roots on the number line
4.   As three interval are formed with roots, check weather each interval satisfies the give quadratic inequality or not 
5.   If it satisfies any interval that will be the solution of the quadratic inequality.