Question

lesson 25 review - solving rational equations
2
\\(\frac{2a^2}{a^2 - 25} - \frac{a}{(a + 5)} = \frac{7}{(a - 5)}\\)
lcd = (a + 5)(a - 5)
\\(\frac{2a^2}{(a + 5)(a - 5)} - \frac{a}{(a + 5)} \frac{(a - 5)}{(a - 5)} = \frac{7}{(a - 5)} \frac{(a + 5)}{(a + 5)}\\)
\\(2a^2 - a(a - 5) = 7(a + 5)\\)
\\(2a^2 - a^2 + 5a = 7a + 35\\)
\\(a^2 + 5a = 7a + 35\\)
\\(a^2 - 2a = 35\\)
\\(a^2 - 2a - 35 = 0\\)
\\((a - 7)(a + 5) = 0\\)
\\(a - 7 = 0\\) or \\(a + 5 = 0\\)
\\(a = 7\\) or \\(a = -5\\) (extraneous root/solution)
steps:

1. factor denominators
2. find lcd
3. multiply to create lcd
4. solve using numerator
5. check for extraneous solutions!

multiple-choice question
listen to this voice note and then answer the following question below:
why is she labeling \\(a = -5\\) as extraneous?

• because it was an extra solution that had to be rejected since it made the lcd negative.
• because it was an extra solution that had to be rejected since it made the lcd equal to 0.

Explanation:

To determine why \( a = -5 \) is extraneous, we analyze the original rational equation. The denominators are \( a^2 - 25=(a + 5)(a - 5) \), \( a + 5 \), and \( a - 5 \). A value that makes any denominator zero is not in the domain. When \( a=-5 \), the denominators \( a + 5 \) and \( a^2 - 25 \) become zero (since \( (-5)+5 = 0 \) and \( (-5)^2-25=25 - 25 = 0 \)). When we solve rational equations, we multiply by the LCD \((a + 5)(a - 5)\), but values that make the LCD (or original denominators) zero must be excluded. So \( a=-5 \) is extraneous because it makes the LCD (and original denominators) equal to 0, violating the domain of the original equation. The first option's reason about "making the LCD negative" is incorrect; the issue is making the LCD zero, not negative. The second option correctly states it's rejected because it makes the LCD (and denominators) equal to 0.

Answer:

B. Because it was an extra solution that had to be rejected since it made the lcd equal to 0.