Question

Which solution to the equation StartFraction 3 Over a + 2 EndFraction + StartFraction 2 Over a EndFraction = StartFraction 4 a minus 4 Over a squared minus 4 EndFraction is extraneous?
a = –2
a = –2 and a = 4
neither a = –2 nor a = 4
a = 4

Answer 1

To determine whether a solution to the equation \(\frac{3}{a+2} + \frac{2}{a} = \frac{4a - 4}{a^2 - 4}\) is extraneous, we first need to find the solutions for \(a\) and then test them.

Given the equation:
\[
\frac{3}{a+2} + \frac{2}{a} = \frac{4a - 4}{a^2 - 4}
\]

Notice that \(a^2 - 4\) can be factored:
\[
a^2 - 4 = (a - 2)(a + 2)
\]

Now, rewrite the right-hand side:
\[
\frac{4a - 4}{(a-2)(a+2)} = \frac{4(a - 1)}{(a-2)(a+2)}
\]

This tells us that the denominator is zero when \(a = 2\) or \(a = -2\). Hence, \(a = -2\) is a value to check, as it makes the left-side undefined.

Next, let's find the common denominator for the left side:
The common denominator for \(\frac{3}{a+2}\) and \(\frac{2}{a}\) is \(a(a + 2)\).

Rewriting the left side:
\[
\frac{3a}{a(a + 2)} + \frac{2(a + 2)}{a(a + 2)} = \frac{3a + 2a + 4}{a(a + 2)} = \frac{5a + 4}{a(a + 2)}
\]

Now, we equate both sides:
\[
\frac{5a + 4}{a(a + 2)} = \frac{4(a - 1)}{(a - 2)(a + 2)}
\]

Cross-multiplying gives:
\[
(5a + 4)(a - 2) = 4(a - 1)a
\]

Expanding both sides:
\[
5a^2 - 10a + 4a - 8 = 4a^2 - 4a
\]
\[
5a^2 - 6a - 8 = 4a^2 - 4a
\]
\[
5a^2 - 4a^2 - 6a + 4a - 8 = 0
\]
\[
a^2 - 2a - 8 = 0
\]

Factoring:
\[
(a - 4)(a + 2) = 0
\]

So, we have potential solutions:
\[
a = 4 \quad \text{or} \quad a = -2
\]

Now let's check if either solution makes the original equation invalid due to division by zero:
- For \(a = -2\), the denominators in the original equation lead to undefined expressions (specifically \(\frac{3}{0}\)). Therefore, \(a = -2\) is an extraneous solution.
- For \(a = 4\), it does not make any denominator zero, so it is a valid solution.

Thus the answer to the question of which solution is extraneous is:
\[
\text{a = -2}
\]