Choose the correct answer from the given four options in the following questions:
$ \left(x^{2}+1\right)^{2}-x^{2}=0 $ has
(A) four real roots
(B) two real roots
(C) no real roots
(D) one real root.

To do:

We have to find the correct answer.

Solution:

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

$x^{4}+1+2 x^{2}-x^{2}=0$

$x^{4}+x^{2}+1=0$

Let $x^{2}=k$

This implies,

$(x^{2})^{2}+x^{2}+1=0$

$k^{2}+k+1=0$

Comparing with $ak^{2}+bk+c=0$, we get,

$a =1, b=1$ and $c=1$

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

$=(1)^{2}-4(1)(1)$

$=1-4$

$=-3<0$

Therefore,

$k^{2}+k+1=0$

$x^{4}+x^{2}+1=0$

Hence, $(x^{2}+1)^{2}-x^{2}=0$ has no real roots.

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Updated on: 10-Oct-2022

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