If frac{x^{2}}{x-5}

Question

(A) Given the inequality $\frac{x^{2}}{x-5} < 0$.
Since $x^2 \ge 0$ for all real $x$,the expression $\frac{x^2}{x-5}$ is negative only when the de

Vedclass · Accepted Answer

$(-\infty, 0) \cup (0, 5)$

Vedclass · Answer

(A) Given the inequality $\frac{x^{2}}{x-5} < 0$.
Since $x^2 \ge 0$ for all real $x$,the expression $\frac{x^2}{x-5}$ is negative only when the denominator is negative and the numerator is non-zero.
$1$. The numerator $x^2 > 0$ implies $x 
eq 0$.
$2$. The denominator $x-5 < 0$ implies $x < 5$.
Combining these conditions,we get $x < 5$ and $x 
eq 0$.
Thus,the solution set is $x \in (-\infty, 0) \cup (0, 5)$.

If $\frac{x^{2}}{x-5} < 0$,then $x \in$

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