ax

A real number α is called the root or solution of the quadratic equation if

aα

- The root of the quadratic equation is the zeroes of the polynomial p(x).
- We know from chapter two that a polynomial of degree can have max two zeroes. So a quadratic equation can have maximum two roots
- A quadratic equation has no real roots if b
^{2}- 4ac < 0

S.no |
Method |
Working |

1 |
factorization |
This method we factorize the equation by splitting the middle term b In ax ^{2}+bx+c=0Example 6x ^{2}-x-2=01) First we need to multiple the coefficient a and c.In this case =6X-2=-12 2) Splitting the middle term so that multiplication is 12 and difference is the coefficient b 6x ^{2} +3x-4x-2=03x( 2x+1) -2(2x+1)=0 (3x-2) (2x+1)=0 3) Roots of the equation can be find equating the factors to zero 3x-2=0 => x=3/2 2x+1=0 => x=-1/2 |

2 |
Square method |
In this method we create square on LHS and RHS and then find the value. ax ^{2} +bx+c=01) x ^{2} +(b/a) x+(c/a)=02) ( x+b/2a) ^{2} –(b/2a)^{2} +(c/a)=03) ( x+b/2a) ^{2}=(b^{2}-4ac)/4a^{2}4) Example x ^{2} +4x-5=01) (x+2) ^{2} -4-5=02) (x+2) ^{2}=93) Roots of the equation can be find using square root on both the sides x+2 =-3 => x=-5 x+2=3=> x=1 |

3 |
Quadratic method |
For quadratic equation ax ^{2} +bx+c=0,roots are given by $x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$ and $x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$ For b ^{2} -4ac > 0, Quadratic equation has two real roots of different valueFor b ^{2}-4ac =0, quadratic equation has one real rootFor b ^{2}-4ac < 0, no real roots for quadratic equation |

S.no |
Condition |
Nature of roots |

1 |
b^{2} -4ac > 0 |
Two distinct real roots |

2 |
b^{2}-4ac =0 |
One real root |

3 |
b^{2}-4ac < 0 |
No real roots |

x

Solution

There is no constant term in this quadratic equation, we can x as common factor

x(x-6)=0

So roots are x=0 and x=6

2) Solve the quadratic equation

x

Solution

x

x

or x=4 or -4

3) Solve the quadratic equation by factorization method

x

Solution

1) First we need to multiple the coefficient a and c.In this case =1X-20=-20

The possible multiple are 4,5 ,2,10

2) The multiple 4,5 suite the equation

x

x(x-5)+4(x-5)=0

(x+4)(x-5)=0

or x=-4 or 5

4) Solve the quadratic equation by Quadratic method

x

Solution

For quadratic equation

ax

roots are given by

$x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}$

and

$x=\frac{-b-\sqrt{b^{2}-4ac}}{2a}$

Here a=1

b=-3

c=-18

Substituting these values,we get

x=6 and -3

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