find the roots of the following quadratic equations by factorization

we shall study quadratic equations and solve them by various methods. In this topic, we will find the roots of the following quadratic equations by factorization method.

Roots of Quadratic Equation Class 10 

Now we will discuss how to solve the following quadratic equation by factorization method

Example -1 

Find the roots of the quadratic equation `x^2-9x+18=0` by factorization method.

Solution 

The given equation is  `x^2-9x+18=0` 

`\Rightarrow\ \ \ \ x^2-(6+3)x\ +\ 18\ =0` 

`\Rightarrow\ \ \ \ x^2-\ \ 6x-3x\ +\ 18\ =0` 

`\Rightarrow\ \ \ \ x(x-\ 6)-3(x\ -\ 6)\ =0` 

`\Rightarrow\ \ \ \ (x-\ 6)(x\ -\ 3)\ =0`

`\Rightarrow\ \ \ \ x-\ 6=\ 0\ or\ x\ -\ 3\ =0` 

`\Rightarrow\ \ \ \ x=\ 6\ \ \ \ or\ \ \ x=3` 

Hence, `x=\ 6\ \ \ \ or\ \ \ x=3` are the roots of the given quadratic equation.

Example -2 

Find the roots of the quadratic equation `2x^2-5x+3=0`  by factorization method.

Solution 

The given equation is  `2x^2-5x+3=0` 

`\Rightarrow\ \ 2x^2\ -\ (2+3)x+ 3 =0`

`\Rightarrow\ \ 2x^2\ -\ 2x-3x+3=0`

`\Rightarrow\ \ \ \ 2x(x-1)-3(x\ -\ 1)=0`

`\Rightarrow\ \ \ \ (x-1)(2x\ -\ 3)=0` 

`\Rightarrow\ \ \ \ x-1=0\ \ \ \ or\ \ \ 2x\ -\ 3=0` 

`\Rightarrow\ \ \ \ \ x\ \ =1\ \ or\ x\ =\ \frac{3}{2}` 

Hence, `x\ \ =1\ \ or\ x\ =\ \frac{3}{2}`  are the toots of given quadratic equation.

Example -3 

Find the roots of the quadratic equation `6x^2-x\ -\ 2\ =0`  by factorization method.

Solution 

The given equation is  `6x^2-x\ -\ 2\ =0`  

`\Rightarrow\ \ \ 6x^2\ -(4\ -\ 3)x\ \ \ -\ 2\ =0` 

`\Rightarrow\ \ \ 6x^2\ -4x\ \ +\ 3x\ \ \ -\ 2\ =0`

`\Rightarrow\ \ \ 2x(3x\ -\ 2)+\ \ 1(3x\ \ -\ 2)\ =0` 

`\Rightarrow\ \ \ (3x\ -\ 2)+\ (2x\ \ +1\ )\ =0` 

`\Rightarrow\ \ \ 3x\ -\ 2=0\ \ \ \ or\ \ 2x\ \ +1\ \ =0` 

⟹   `x =\frac {2}{3}`      or  `x  =  -\frac {1}{2}`    

Hence,  `x =\frac {2}{3}`      or  `x  =  -\frac {1}{2}`   are the roots of given quadratic equation.

Example -4

Find the roots of the quadratic equation `3x^2- 2\sqrt6x+2 =0` by factorization method.

Solution

 The given equation is `3x^2- 2\sqrt6x+2 =0`  

⟹    `3x^2-(\sqrt6+\sqrt6)x+2=0`               

 ⟹    `3x^2-\sqrt6x+\sqrt6x+ 2=0` 

 

⟹  `\sqrt3x(\sqrt3x-\sqrt2) - \sqrt2(\sqrt3x- \sqrt2)=0`

  

⟹ `(\sqrt3x-\sqrt2) (\sqrt3x -\sqrt2)=0`

⟹ `\sqrt3x -\sqrt2 =0`  or `\sqrt3x - \sqrt2 =0`

 

⟹      `x=\sqrt\frac {2}{3}`   or        `x=\sqrt\frac {2}{3}`

  

Hence,   `x=\sqrt\frac {2}{3}`   or     `x=\sqrt\frac {2}{3}`     are the roots of the given quadratic equation.

 Example -5 

Find the roots of the quadratic equation `15a^2 x^2-16abx-15b^2=0`   by factorization method.

Solution 

The given equation is   `15a^2 x^2-16abx-15b^2=0` 

⟹  `15a^2 x^2-(25ab-9ab)x-15b^2=0`

⟹  `15a^2 x^2-25abx+9abx-15b^2=0` 

⟹  `5ax(3ax-5b)+3b(3ax-5b)=0`

⟹  `(3ax-5b)(5ax+3b)=0` 

⟹      `3ax-5b=0 or 5ax+ 3b=0` 

⟹      `x=\frac {5b}{3a}`     or `x= -\frac{3b}{5a}=0` 

Hence,  `x=\frac {5b}{3a}`     or `x= -\frac{3b}{5a}=0` 

 are the roots of the given quadratic equation.

Now you must have known the method of finding roots of quadratic equation by factoring.