The number of real roots of the equation frac | vedclass

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Question

(B) Given equation is $\frac{(x^2+1)^3}{x^3} + \frac{x^2+1}{3x} = 0$ for $x 
eq 0$. Let $t = \frac{x^2+1}{x}$. Then the equation becomes $t^3 + \frac

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given equation is $\frac{(x^2+1)^3}{x^3} + \frac{x^2+1}{3x} = 0$ for $x 
eq 0$. Let $t = \frac{x^2+1}{x}$. Then the equation becomes $t^3 + \frac{t}{3} = 0$. Factoring out $t$,we get $t(t^2 + \frac{1}{3}) = 0$. This implies $t = 0$ or $t^2 = -\frac{1}{3}$. Since $t$ must be a real number,$t^2 = -\frac{1}{3}$ has no real solutions. Thus,we must have $t = 0$,which means $\frac{x^2+1}{x} = 0$. This implies $x^2 + 1 = 0$,or $x^2 = -1$. Since $x^2 = -1$ has no real solutions for $x$,there are no real roots for the given equation. Therefore,the number of real roots is $0$.

The number of real roots of the equation $\frac{(x^2+1)^3}{x^3} + \frac{x^2+1}{3x} = 0, (x \neq 0)$ is

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