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Question

(B) Given equation: $x^{2}+\frac{1}{x^{2}}=-2$ $(x 
eq 0)$Multiply the entire equation by $x^{2}$:$x^{2}(x^{2}) + x^{2}(\frac{1}{x^{2}}) = -2(x^{2})$

Vedclass · Accepted Answer

No,it is not a quadratic equation.

Vedclass · Answer

(B) Given equation: $x^{2}+\frac{1}{x^{2}}=-2$ $(x 
eq 0)$Multiply the entire equation by $x^{2}$:$x^{2}(x^{2}) + x^{2}(\frac{1}{x^{2}}) = -2(x^{2})$$x^{4} + 1 = -2x^{2}$Rearrange the terms to form a polynomial equation:$x^{4} + 2x^{2} + 1 = 0$Let $p(x) = x^{4} + 2x^{2} + 1$.The degree of this polynomial is $4$.$A$ quadratic equation must have a degree of $2$.Since the degree is $4$,the given equation is not a quadratic equation.

Examine whether the following equation is quadratic or not: $x^{2}+\frac{1}{x^{2}}=-2$ $(x \neq 0)$

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