Question

Find the roots of the following quadratic equation, if they exist, using the quadratic formula of Shridhar Acharya.
$2x^2-2\sqrt{2}x+1=0$.

Answer 1

For the equation $2x^2-2\sqrt{2}x+1=0;a=2,b=-2\sqrt{2},c=1$

By Shridhar acharya's formula to test the existence of the roots,

we have $b^2-4ac=(2\sqrt{2}{)}^2-4\times 2\times 1$

$=8-8=0$

We got $\Delta =0$

$\therefore$ Root for the equation exists and they are equal.

Let the roots of the equation be $x_1$ and $x_2$

Then $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$

$\Rightarrow x_1=\frac{2\sqrt{2}+0}{2\times 2}=\frac{1}{\sqrt{2}}$

$\Rightarrow x_2=\frac{2\sqrt{2}+0}{2\times 2}=\frac{1}{\sqrt{2}}$