Solve the equation, eliminating any extraneous solutions.

[tex]\[ \frac{x^2 + 5}{x + 5} = \frac{30}{x + 5} \][/tex]

A. [tex]\(x = -25\)[/tex]

B. [tex]\(x = 5\)[/tex]

C. [tex]\(x = -5, 5\)[/tex]

D. [tex]\(x = -5\)[/tex]



Answer :

Certainly! Let's solve the equation [tex]\(\frac{x^2 + 5}{x + 5} = \frac{30}{x + 5}\)[/tex] step-by-step and eliminate any extraneous solutions.

1. Start with the given equation:

[tex]\[ \frac{x^2 + 5}{x + 5} = \frac{30}{x + 5} \][/tex]

2. Since both sides of the equation have the common denominator [tex]\((x + 5)\)[/tex], first consider the scenario where [tex]\(x \neq -5\)[/tex] to avoid division by zero:

[tex]\[ (x^2 + 5) = 30 \][/tex]

3. Solving the simplified equation:

[tex]\[ x^2 + 5 = 30 \][/tex]

Subtract 5 from both sides:

[tex]\[ x^2 = 25 \][/tex]

4. Take the square root of both sides:

[tex]\[ x = \pm 5 \][/tex]

Hence, the potential solutions are [tex]\(x = 5\)[/tex] and [tex]\(x = -5\)[/tex].

5. Check for extraneous solutions:

Since we earlier excluded [tex]\(x = -5\)[/tex] to avoid division by zero (since [tex]\(\frac{x^2 + 5}{x + 5}\)[/tex] and [tex]\(\frac{30}{x + 5}\)[/tex] would be undefined at [tex]\(x = -5\)[/tex]), we discard [tex]\(x = -5\)[/tex] as an invalid solution.

6. Confirm the valid solution:

We are left with [tex]\(x = 5\)[/tex].

So, the only valid solution to the equation is:

[tex]\[ x = 5 \][/tex]