Question

find the roots of the equation 5x^2-6x-12=0 by the method of completing the square​

Answer 1

$\bold{5x^2-6x-12=0}$

Dividing by 5,

$\implies\bold{\frac{5x^2-6x-12}{5}=\frac{0}{5}}$

$\implies\bold{\frac{5x^2}{5}-\frac{6x}{5}-\frac{12}{5}=0}$

$\implies\bold{x^2-\frac{6x}{5}-\frac{12}{5}=0}$

We know that ,

$\boxed{\bold{(a-b)^2=a^2-2ab+b^2}}$

Here,

a = x

$\bold{-2ab=\frac{-6x}{5}}$

$\bold{-2xb=\frac{-6x}{5}}$       (as a = x)

$\bold{-2b=\frac{-6}{5}}$

$\implies\bold{b=\frac{-6}{5\times(-2)}}$

$\implies\bold{b=\frac{-6}{-10}}$

$\boxed{\bold{\therefore{b=\frac{3}{5}}}}$

Now in our equation,

$\bold{x^2-\frac{6x}{5}-\frac{12}{5}=0}$

Adding and subtracting $\bold{(\frac{3}{5})^2}$

$\implies\bold{x^2-\frac{6x}{5}-\frac{12}{5}+(\frac{3}{5})^2-(\frac{3}{5})^2=0}$

$\implies\bold{x^2-\frac{6x}{5}+(\frac{3}{5})^2-\frac{12}{5}-(\frac{3}{5})^2=0}$

$\implies\bold{(x-\frac{3}{5})^2-\frac{12}{5}-(\frac{3}{5})^2=0}$

$\implies\bold{(x-\frac{3}{5})^2=\frac{12}{5}+(\frac{3}{5})^2}$

$\implies\bold{(x-\frac{3}{5})^2=\frac{12}{5}+\frac{9}{25}}$

$\implies\bold{(x-\frac{3}{5})^2=\frac{9+12(5)}{25}}$

$\implies\bold{(x-\frac{3}{5})^2=\frac{9+60}{25}}$

$\implies\bold{(x-\frac{3}{5})^2=\frac{69}{25}}$

$\implies\bold{(x-\frac{3}{5})^2=\frac{(\sqrt69)^2}{(5)^2}}$

Now cancelling the squares on both sides,

$\implies\bold{(x-\frac{3}{5}) =\±\frac{\sqrt69}{5}}$

Solving,

$\bold{(x-\frac{3}{5}) =+\frac{\sqrt69}{5}}$                     $\bold{(x-\frac{3}{5}) =-\frac{\sqrt69}{5}}$

$\bold{x =\frac{\sqrt69}{5}+\frac{3}{5}}$                           $\bold{x =-\frac{\sqrt69}{5}+\frac{3}{5}}$

$\bold{x =\frac{\sqrt69+3}{5}}$                              $\bold{x =\frac{-\sqrt69+3}{5}}$

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So, the roots of the equation are :

$\boxed{\bold{x =\frac{\sqrt69+3}{5}}}$              $\boxed{\bold{x =\frac{-\sqrt69+3}{5}}}$