In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

$2x^2-2\sqrt2 x+1=0$

Given:

Given quadratic equation is $2x^2-2\sqrt2 x+1=0$.

To do:

We have to determine whether the given quadratic equation has real roots.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=2, b=-2\sqrt2$ and $c=1$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

Therefore,

$D=(-2\sqrt2)^2-4(2)(1)=4(2)-8=8-8=0$.

As $D=0$, the given quadratic equation has real and equal roots and the roots are

$x=\frac{-b\pm \sqrt{D}}{2a}$

$x=\frac{-(-2\sqrt2)\pm \sqrt{0}}{2(2)}$ 

$x=\frac{2\sqrt2}{4}$ 

$x=\frac{\sqrt2}{2}$ 

$x=\frac{\sqrt2}{\sqrt2\times\sqrt2}$

$x=\frac{1}{\sqrt2}$

The roots are $\frac{1}{\sqrt2}$ and $\frac{1}{\sqrt2}$.  

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Updated on: 10-Oct-2022

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