Solve x/(x + 5) − 25/(x² + 25) = 0.

Step-by-step solution

1. 1

   Note the domain first. The denominator $x^{2}+25$ is never zero for real $x$, so the only excluded value is $x=-5$ from the first fraction. Any candidate root equal to $-5$ would have to be rejected.

2. 2

   Move one fraction across and cross-multiply. The equation says the two fractions are equal:

   $\frac{x}{x+5}=\frac{25}{x^{2}+25}\quad\Longrightarrow\quad x\left(x^{2}+25\right)=25(x+5)$

3. 3

   Expand — and watch the middle terms cancel.

   $x^{3}+25x=25x+125\quad\Longrightarrow\quad x^{3}=125$

   The $25x$ on each side cancels exactly, which is what turns an apparently awkward cubic into a bare cube root.

4. 4

   Take the real cube root.

   $x=\sqrt[3]{125}=5$

   Over the reals this is the only solution; the cubic $x^{3}-125=(x-5)\left(x^{2}+5x+25\right)$ has a quadratic factor with discriminant $25-100=-75<0$, so the other two roots are complex.

5. 5

   Check against the original equation and the domain. $x=5\neq -5$, so it is admissible, and

   $\frac{5}{10}-\frac{25}{50}=0.5-0.5=0\ \checkmark$

   $\boxed{x=5}$

6. 6

   Watch out for the common misreading. If the second denominator were $x^{2}-25$ instead of $x^{2}+25$, the problem would be entirely different — $x^{2}-25$ factors as $(x+5)(x-5)$ and brings a second excluded value $x=5$, which would then have to be rejected. Always confirm the sign inside the denominator before clearing fractions.