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(A) Given equation is $(x^{2}+1)^{2}-x^{2}=0$.Expanding the square: $x^{4}+1+2x^{2}-x^{2}=0$,which simplifies to $x^{4}+x^{2}+1=0$.Let $x^{2}=y$. Sinc

Vedclass · Accepted Answer

no real roots

Vedclass · Answer

(A) Given equation is $(x^{2}+1)^{2}-x^{2}=0$.Expanding the square: $x^{4}+1+2x^{2}-x^{2}=0$,which simplifies to $x^{4}+x^{2}+1=0$.Let $x^{2}=y$. Since $x$ is a real number,$x^{2} \ge 0$,so $y \ge 0$.The equation becomes $y^{2}+y+1=0$.Comparing with $ay^{2}+by+c=0$,we have $a=1, b=1, c=1$.The discriminant $D = b^{2}-4ac = (1)^{2}-4(1)(1) = 1-4 = -3$.Since $D Therefore,the original equation $(x^{2}+1)^{2}-x^{2}=0$ has no real roots.

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

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