## HOW TO FIND THE SOLUTIONS OF QUADRATIC EQUATIONS

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Today we are going to discuss Quadratic Equations , different methods of solution of Quadratic equation , its roots , Discriminant  and its formation with the help of some examples.

## Any expression of the form   ax2 + bx + c = 0,

where x represents a variable , and ab, and c are constants but a ≠ 0 , is known as Quadratic Equation.The numbers ab, and c are called  coefficients   of the equation.
ax2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

Examples
1   x2 + 4x − 4 = 0
2   3x2 + 4x +4 = 0
3   2x2 - 4x − 5 = 0
4   5x2 -7x + 3 = 0
5   10 x2 + 9x − 8= 0
6   1.5x2 + 4.8x − 4.3 = 0
7   3x2 + 2x + 1 = 0
8  x2 +5x −20 = 0
9  2x2  – 3x + 1 = 0
10  4x – 3x2  + 2 = 0 and
11  1 – x2 + 300x = 0

## Discriminant

The number D = b2 – 4ac determined from the coefficients of the equation ax2 + bx + c = 0 is called the Discriminant of the Quadratic Equation.
Let us consider

x2  +5x −20 = 0, then a = 1, b = 5 and c = - 20,
D = b2 – 4ac = 52 – 4×1×(-20) = 25 + 80 = 105

2x2 + 4x − 4 = 0,  a = 2,b = 4 and c = -4
D = b2 – 4ac = 42 – 4×2×(-4) = 16+32 = 48

x2 + 4x + 4 = 0 , a = 1,b = 4 and c = 4
D = b2 – 4ac = 42 – 4×1×(4) = 16-16 = 0

3x2 + 6x + 4 = 0 , a=3,b=6 and c= 4
D = b2 – 4ac = 62 – 4×3×(4) = 36 - 48 = -12

## What are  Roots of Quadratic Equation

By roots of the quadratic Equation we means that values which we put in given equation satisfies the given quadratic Equation.

For example in this equation 5x2 +20 x - 25 = 0 if we put values "1" and "5" in place of x then these values shall be called the roots of this equation ,Because these values satisfy the given equation.

Similarly the equation 4x2 - 16 = 0 shall have roots 4 and -4 ,because they shall satisfy the given equation

## Nature of Roots

The nature of roots of Quadratic Equation can be discussed with the help of Discriminant .

1 When D = 0 Then roots are real and Equal, It means the roots of quadratic equation are without √  and both the roots are equal to each other . Some examples are  (5,5) , (6,6) , (8,8) etc

2 When D <0 Then roots are not real But they are complex conjugate of each other. It means the roots of given equation have roots like √ (-) term in it. Some examples are 2+ √ (-5)  and 2 - √ (-5)

3 When D >0 Then roots are real and Unequal

a> When D is Positive but not perfect square then roots exists in Quadratic Surd . Some of the examples are 3+√ 5 and 3-√ 5

b> When D is Positive and perfect square then roots are rational provided a,b and c are rational. Some example are 3 and 2

## Note:

If l,m are two roots of any quadratic Equation ax2 + bx + c = 0  ,then
Sum of roots     l + m  = - b/a

Product of roots   l × m = c/a

If  sum and the product of roots of any equation is given then, that equation can be written as follows
x2 + ( sum of the roots ) x + Product of roots = 0
x2 + ( -b/a ) x + (c/a) = 0
or
x2 +  ( l+m ) x  +  lm = 0

1  If sum of all the coefficients of a quadratic equation is equal to Zero ,then one of the root of that  equation is 1 i.e. “ unity”.

## 5 Any quadratic Equation will have equal in magnitude but opposite in  sign roots , if   coefficients of x equal to zero . i.e. b = 0.

### 2 Completing the Square

Factorisation Method

To  solve  ax2 + bx + c = 0 by Factorisation Method

1 Split the middle term into two terms in such a way that their         product must be equal to the product of a and c.

### 3  Take whichever  is common in the two factors

4  Repeat the process of step 3

5  Put Both Factors equal  to Zero and calculate the values of x

### Example

4x2 + 4x - 3 = 0

4x2  -2x +6x - 3 = 0  Split the middle term into two terms

2x (2x-1) +3( 2x - 1) = 0     Follow  Step 2

(2x-1)(2x+3) = 0                 Follow  Step 2

Either  2x - 1 = 0  or     2x + 3=0
x = 1/2   or      x = -3/2

### Example 2x2 – 5x + 3 = 0, 2x2 – 2x – 3x + 3=0   ( Split the middle term into two terms ) 2x (x – 1) –3(x – 1) = 0   ( Taking whichever is common ) (2x – 3)(x – 1)=0 Either  2x - 3 =  0  or x – 1 = 0  Now, 2x = 3 and   x = 1. x=3/2   and   x = 1. Example 10x2 + 21 x - 10 = 0  10x2 +25x - 4x - 10 = 0   ( Split the middle term into two terms ) 5x (2x + 5) -2 ( 2x + 5) = 0  Follow  Step 2  (2x+5)(5x-2) = 0    Follow  Step 2  Either  2x + 5 = 0  or     5x - 2=0                x = -5/2   or      x= 2/5

Completing the Square

1 Shift the constant term to right hand side of equal sign.

2 Complete the square in left side and add the term which is missing and adjust the added term on the right side.

3 Equate the Left hand term to Right hand term.

Let us consider a Quadratic Equation

9x2 – 15x + 6 = 0
To make the complete square add the missing term   (b)2  and  subtract the same term

(3x)2  – 2*3x *(5/2)+ (5/2)2–  (5/2)2 +6 = 0

{3x-(5/2)}2 -(25/4)+6 = 0

{3x - (5/2)}2 -(1/4) = 0

{3x-(5/2)}2  = (1/4)                      Taking square roots
3x - (5/2) = (1/2)  or  3x-(5/2) = - (1/2)
3x = (1/2)+(5/2)  or 3x = -(1/2) + (5/2)
3x = 6/2 or 3x = 4/2
x = 1 or    3x = 2
x = 1 or    x = 2/3

The roots of the given equation are 1 and   2/3

### Example

9x2 + 12 x - 1 = 0
9x2 + 12 x = 1
(3x)2 + 2× 3x× 2 + 22 = 1+22

(3x+2)2 =1+4

(3x+2)2 =5      Taking Square Roots

3x+2 =  5                    or   3x+2 = - ( 5)
3 x =  5  -   2                or     3x = - ( 5) -2
x = (  5  -  2  )/3      or   x = (  -  5 - 2 ) /3

Roots are real and Unequal

Example

4x2 + 16 x - 9 = 0
4x2 + 16 x = 9
(2x)2 + 2× 2x× (4) + (4)2 = 9+(4)2
(2x+4)2 =9+16
(2x+4)2 =25
(2x+4)2 =52   Taking Square Roots
2x+4 = 5
2x = 5-4
x = 1/2
Two  real and Equal roots

Example

x2 -6 x + 11 = 0
x2 -6 x = - 11
(x)2 -2(x)(3)+ 32  = -11 + 9

(x-3)2 = -11+9

(x-3)2 = -2

As the Square of any real number can not be negative , so this
equation have NO Real Roots , The equation have only complex
roots.

### Example

4x2 -40 x +100 = 0

Since  D = b2 -4ac
D(-40)2 - 4×4×100
D = 1600-1600
D= 0

x= -b/2a as second part of the formula vanishes i. e .

x = -(-40)/(2×4)
x  = 40/8
x = 5
Here both roots are equal and Real

Example

4x2 - 20 x +29 = 0

Since  D = b2 -4ac

D = (-20)2 - 4×4×29

D= 400 - 464= -64

The Squar root of -64 is  -8i

Since Discriminant is negative , so this quadratic Equation have no real roots but two complex roots can be found.

x= {-b+8 i} /(2a)    and  {-b-8 i}/{1/2(a)}

x = {-(-20)+8 i}/{2×4} and {-(-20)-8 i} /{2×4}
x={20+8i}/8  and    {20 - 8 i}/{8}

These are two roots of Quadratic Equation which are complex conjugate of each other .

Example

6x2-6x -1=0

Using the Quadratic Equation formula for the variable $\displaystyle{r}$:

$\displaystyle=\frac{{{6}\pm\sqrt{{60}}}}{{12}}$
$\displaystyle=\frac{{{6}-{2}\sqrt{{15}}}}{{12}}{\quad\text{or}\quad}\frac{{{6}+{2}\sqrt{{15}}}}{{12}}$
$\displaystyle=\frac{{{3}-\sqrt{{15}}}}{{6}}{\quad\text{or}\quad}\frac{{{3}+\sqrt{{15}}}}{{6}}$

Hence here two roots are Quadratic surd ( Irrational )

### Example2x2-4x -5=0

Here a = 2 , b = -4 and c = - 5

Using the Quadratic Equation formula for the x

$\displaystyle{x}=\frac{{-{b}\pm\sqrt{{{b}^{2}-{4}{a}{c}}}}}{{{2}{a}}}$

=44± 16+24
$\displaystyle=\frac{{{4}\pm\sqrt{{40}}}}{{4}}$
$\displaystyle=\frac{{{4}-\sqrt{{40}}}}{{4}}{\quad\text{or}\quad}\frac{{{4}+\sqrt{{40}}}}{{4}}$

Roots are Irrational NOT Equal

### Example

2x+ 4x + 2 = 0

Here a = 2 , b = 4 and c = 2

Using the formula ,we get

$\displaystyle=\frac{{{4}\pm\sqrt{{{16}+{24}}}}}{{4}$
$\displaystyle=\frac{{{4}\pm\sqrt{{40}}}}{{4}}$
$\displaystyle=\frac{{{4}-\sqrt{{40}}}}{{4}}{\quad\text{or}\quad}\frac{{{4}+\sqrt{{40}}}}{{4}}$

x  = -1    and   -1

Two roots are  Equal and Real

## Conclusion

In this post I have discussed different method of solutions of  Quadratic equation , its roots , Discriminant . If this post helped you little bit, then please share it with your friends and like this post  to boost me and to do better, and also follow me on my Blog .We shell meet in next post till then Bye .

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