State whether the following quadratic equations have two distinct real roots. Justify your answer.
$ (x-\sqrt{2})^{2}-2(x+1)=0 $

Given:

\( (x-\sqrt{2})^{2}-2(x+1)=0 \)

To do:

We have to state whether the given quadratic equations have two distinct real roots.

Solution:

\( (x-\sqrt{2})^{2}-2(x+1)=0 \)

$x^2+(\sqrt2)^2-2(\sqrt2)x-2x-2=0$

$x^2+2-2\sqrt2x-2x-2=0$

$x^2-(2\sqrt2+2)x=0$

Comparing with $a x^{2}+b x+c=0$, we get,

$a =1, b=2(\sqrt2+1)$ and $c=0$

Discriminant $D=b^{2}-4 a c$

$=[2(\sqrt2+1)]^{2}-4(1)(0)$

$=4(\sqrt2+1)^2-4$

$=4[(\sqrt2+1)^2-1]>0$

$D>0$

Hence, the equation \( (x-\sqrt{2})^{2}-2(x+1)=0 \) has two distinct real roots.

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Updated on: 10-Oct-2022

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