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
⟹ `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.
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