HOW TO FIND THE SOLUTIONS OF QUADRATIC EQUATIONS

Hello Friends welcome to My Blog, 

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.

Quadratic Equation 

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

where x represents a variable , and a, b, and c are constants but a ≠ 0 , is known as Quadratic Equation.The numbers a, b, 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

Some Important facts about Quadratic Equation

 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”.

5x² +20 x - 25 = 0  ,  2x² -10 x + 8 = 0

2. when a quadratic equation has one root equal to zero, then it constant term must be equal to zero.

4x² -20 x   = 0 ,  2x² -12 x  = 0 

3. Any quadratic Equation will have reciprocal roots ,if its  coefficients of x² term and constant term are equal  i. e.  a = c.

4x² -20 x +4 = 0 , 2x² -20 x +2 = 0 

4 Any quadratic Equation will have negative reciprocal  roots ,if   coefficients of x² and constant term are equal in magnitude but opposite in sign. i.e  a = - c.

4x² -20 x -4 = 0 ,  2x² -20 x -2 = 0 

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.

4x²  - 16 = 0 , 5x²  -25 = 0 ,  -4x²  -36 = 0 

Solving the Quadratic Equation

1 Factorisation Method

2 Completing the Square

3 Quadratic Formula

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.

2 Take whichever is common in 1st two and Last two  factors.

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 -3 x - 35 = 0 

2x² -10x +7x - 35 = 0 

(  Split the middle term into two terms )

2x (x - 5) +7( x - 5) = 0    Follow  Step 2 

(x-5)(2x+7) = 0                 Follow  Step 2 

Either  x - 5 = 0  or     2x + 7=0

               x = 5   or      x= -7/2

Example

4x² -20 x +25 = 0 

4x² -10x -10x + 25 = 0 
( Split the middle term into two terms )

2x (2x - 5) -5( 2x - 5) = 0   Follow  Step 2 

(2x-5)(2x-5) = 0         Follow  Step 2 

Either  2x - 5 = 0  or     2x -5=0

               x = 5/2   or      x= 5/2

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.

Quadratic Formula

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 $\displaystyle{r}$:

=126±
√  (36+24​​

$\displaystyle=\frac{{{6}\pm\sqrt{{60}}}}{{12}}$

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

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

Hence here two roots are Quadratic surd ( Irrational ) 

Example


2x2-4x -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
​​

$\displaystyle=\frac{{{4}\pm\sqrt{{40}}}}{{4}}$

=44±
√ 40​​

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

=44−
√ 40  or44+√ 40​​

Roots are Irrational NOT Equal

Example

2x2 + 4x + 2 = 0

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

Using the formula ,we get

$\displaystyle=\frac{{{4}\pm\sqrt{{{16}+{24}}}}}{{4}$
$\frac{-\; 4\pm \sqrt{}}{4}$

$\displaystyle=\frac{{{4}\pm\sqrt{{40}}}}{{4}}$

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

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 .

Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .