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Question

(B) Given equation: $x+\frac{1}{x}=x^{2}$Multiply both sides by $x$ (since $x 
eq 0$):$x^{2}+1=x^{3}$Rearranging the terms to one side:$x^{3}-x^{2}-1

Vedclass · Accepted Answer

No,it is not a quadratic equation.

Vedclass · Answer

(B) Given equation: $x+\frac{1}{x}=x^{2}$Multiply both sides by $x$ (since $x 
eq 0$):$x^{2}+1=x^{3}$Rearranging the terms to one side:$x^{3}-x^{2}-1=0$Let $p(x) = x^{3}-x^{2}-1$.The degree of the polynomial $p(x)$ is $3$.$A$ quadratic equation must have a degree of $2$.Since the degree of this equation is $3$,it is a cubic equation,not a quadratic equation.Therefore,$x+\frac{1}{x}=x^{2}$ is not a quadratic equation.

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

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