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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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