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(x^{2}+1)^{2}-x^{2}=0  has

            (A) four real roots                                 (B) two real roots

            (C) no real roots                                    (D) one real root.

Answers (1)

(C) no real roots

Solution

Here the given equation is (x^{2}+1)^{2}-x^{2}=0  

                        (x^{2})^{2}+(1)^{2}+2(x^{2})(1)-x^{2}=0           [using? ]
                     x^{4}+1+2x^{2}-x^{2}=0\\ x^{4}+x^{2}+1=0

            Put       x^{2}=z

                        z^{2}+z+1=0

            Compare with az^{2}+bz+c=0 where a \neq 0

            Here    a=1,b=1,c=1
                       b^{2}-4ac=(1)^{2}-4(1)(1)\\ =1-4=-3\\ b^{2}-4ac<0

            Here the given equation has no real roots.

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