Question
Solve the equation by any method.
StartFraction 4 x Over left parenthesis x minus 4 right parenthesis EndFraction plus StartFraction 16 Over left parenthesis x plus 4 right parenthesis EndFraction equals negative StartFraction 92 Over left parenthesis x squared minus 16 right parenthesis EndFraction
Question content area bottom
Part 1
The solution set is StartSet nothing EndSet
.
(Simplify your answer. Type an exact answer, using radicals and i as needed. Use a comma to separate answers as needed.)
StartFraction 4 x Over left parenthesis x minus 4 right parenthesis EndFraction plus StartFraction 16 Over left parenthesis x plus 4 right parenthesis EndFraction equals negative StartFraction 92 Over left parenthesis x squared minus 16 right parenthesis EndFraction
Question content area bottom
Part 1
The solution set is StartSet nothing EndSet
.
(Simplify your answer. Type an exact answer, using radicals and i as needed. Use a comma to separate answers as needed.)
Answers
GPT-4o mini
To solve the equation
\[
\frac{4x}{x - 4} + \frac{16}{x + 4} = -\frac{92}{x^2 - 16},
\]
we first note that \(x^2 - 16\) can be factored:
\[
x^2 - 16 = (x - 4)(x + 4).
\]
This means we can rewrite the equation as:
\[
\frac{4x}{x - 4} + \frac{16}{x + 4} = -\frac{92}{(x - 4)(x + 4)}.
\]
Next, we can find a common denominator for the left-hand side of the equation, which is \((x - 4)(x + 4)\). Rewrite each term:
1. \(\frac{4x}{x - 4} = \frac{4x(x + 4)}{(x - 4)(x + 4)}\)
2. \(\frac{16}{x + 4} = \frac{16(x - 4)}{(x - 4)(x + 4)}\)
So the left-hand side becomes:
\[
\frac{4x(x + 4) + 16(x - 4)}{(x - 4)(x + 4)}.
\]
Now, simplifying the numerator:
\[
4x(x + 4) + 16(x - 4) = 4x^2 + 16x + 16x - 64 = 4x^2 + 32x - 64.
\]
Now the equation can be rewritten as:
\[
\frac{4x^2 + 32x - 64}{(x - 4)(x + 4)} = -\frac{92}{(x - 4)(x + 4)}.
\]
Since we can multiply both sides by \((x - 4)(x + 4)\) (assuming \(x \neq 4\) and \(x \neq -4\)), we have:
\[
4x^2 + 32x - 64 = -92.
\]
Next, we can add \(92\) to both sides:
\[
4x^2 + 32x - 64 + 92 = 0,
\]
\[
4x^2 + 32x + 28 = 0.
\]
To simplify, we divide the entire equation by 4:
\[
x^2 + 8x + 7 = 0.
\]
Now we will factor the quadratic:
\[
(x + 1)(x + 7) = 0.
\]
Setting each factor to zero gives:
1. \(x + 1 = 0 \implies x = -1\)
2. \(x + 7 = 0 \implies x = -7\)
Now we need to check that these solutions do not make the original denominator zero:
- For \(x = -1\): \(x - 4 \neq 0\) and \(x + 4 \neq 0\).
- For \(x = -7\): \(x - 4 \neq 0\) and \(x + 4 \neq 0\).
Both solutions are valid. Hence, the solution set is:
\[
\{-7, -1\}.
\]
\[
\frac{4x}{x - 4} + \frac{16}{x + 4} = -\frac{92}{x^2 - 16},
\]
we first note that \(x^2 - 16\) can be factored:
\[
x^2 - 16 = (x - 4)(x + 4).
\]
This means we can rewrite the equation as:
\[
\frac{4x}{x - 4} + \frac{16}{x + 4} = -\frac{92}{(x - 4)(x + 4)}.
\]
Next, we can find a common denominator for the left-hand side of the equation, which is \((x - 4)(x + 4)\). Rewrite each term:
1. \(\frac{4x}{x - 4} = \frac{4x(x + 4)}{(x - 4)(x + 4)}\)
2. \(\frac{16}{x + 4} = \frac{16(x - 4)}{(x - 4)(x + 4)}\)
So the left-hand side becomes:
\[
\frac{4x(x + 4) + 16(x - 4)}{(x - 4)(x + 4)}.
\]
Now, simplifying the numerator:
\[
4x(x + 4) + 16(x - 4) = 4x^2 + 16x + 16x - 64 = 4x^2 + 32x - 64.
\]
Now the equation can be rewritten as:
\[
\frac{4x^2 + 32x - 64}{(x - 4)(x + 4)} = -\frac{92}{(x - 4)(x + 4)}.
\]
Since we can multiply both sides by \((x - 4)(x + 4)\) (assuming \(x \neq 4\) and \(x \neq -4\)), we have:
\[
4x^2 + 32x - 64 = -92.
\]
Next, we can add \(92\) to both sides:
\[
4x^2 + 32x - 64 + 92 = 0,
\]
\[
4x^2 + 32x + 28 = 0.
\]
To simplify, we divide the entire equation by 4:
\[
x^2 + 8x + 7 = 0.
\]
Now we will factor the quadratic:
\[
(x + 1)(x + 7) = 0.
\]
Setting each factor to zero gives:
1. \(x + 1 = 0 \implies x = -1\)
2. \(x + 7 = 0 \implies x = -7\)
Now we need to check that these solutions do not make the original denominator zero:
- For \(x = -1\): \(x - 4 \neq 0\) and \(x + 4 \neq 0\).
- For \(x = -7\): \(x - 4 \neq 0\) and \(x + 4 \neq 0\).
Both solutions are valid. Hence, the solution set is:
\[
\{-7, -1\}.
\]