QuestionAugust 24, 2025

Solve the equation. (x)/(x-5)-(5)/(x+5)=(50)/(x^2)-25 pm 5i varnothing pm 5 (1)/(5)

Solve the equation. (x)/(x-5)-(5)/(x+5)=(50)/(x^2)-25 pm 5i varnothing pm 5 (1)/(5)
Solve the equation.
(x)/(x-5)-(5)/(x+5)=(50)/(x^2)-25
 pm 5i 
varnothing 
 pm 5 
 (1)/(5)

Solution
4.2(278 votes)

Answer

\varnothing Explanation 1. Simplify the Right Side Recognize x^2 - 25 = (x-5)(x+5). Thus, \frac{50}{x^2 - 25} = \frac{50}{(x-5)(x+5)}. 2. Combine Left Side Rewrite as \frac{x(x+5) - 5(x-5)}{(x-5)(x+5)} = \frac{x^2 + 5x - 5x + 25}{(x-5)(x+5)} = \frac{x^2 + 25}{(x-5)(x+5)}. 3. Equate and Solve Set \frac{x^2 + 25}{(x-5)(x+5)} = \frac{50}{(x-5)(x+5)}. Cancel denominators to get x^2 + 25 = 50. 4. Solve Quadratic Equation x^2 = 25 \implies x = \pm 5. 5. Check for Extraneous Solutions x = 5 or x = -5 makes the original denominators zero. Therefore, they are not valid solutions.

Explanation

1. Simplify the Right Side<br /> Recognize $x^2 - 25 = (x-5)(x+5)$. Thus, $\frac{50}{x^2 - 25} = \frac{50}{(x-5)(x+5)}$.<br />2. Combine Left Side<br /> Rewrite as $\frac{x(x+5) - 5(x-5)}{(x-5)(x+5)} = \frac{x^2 + 5x - 5x + 25}{(x-5)(x+5)} = \frac{x^2 + 25}{(x-5)(x+5)}$.<br />3. Equate and Solve<br /> Set $\frac{x^2 + 25}{(x-5)(x+5)} = \frac{50}{(x-5)(x+5)}$. Cancel denominators to get $x^2 + 25 = 50$.<br />4. Solve Quadratic Equation<br /> $x^2 = 25 \implies x = \pm 5$.<br />5. Check for Extraneous Solutions<br /> $x = 5$ or $x = -5$ makes the original denominators zero. Therefore, they are not valid solutions.
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