Question

lesson 2.5 review - solving rational equations
② (\frac{2a^2}{a^2 - 25} - \frac{a}{(a + 5)} = \frac{7}{(a - 5)}) (\text{lcd} = (a + 5)(a - 5))
(\frac{2a^2}{(a + 5)(a - 5)} - \frac{a}{(a + 5)} cdot \frac{(a - 5)}{(a - 5)} = \frac{7}{(a - 5)} cdot \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) or (a = -5) (rejected)
① factor denominators
② find lcd
③ multiply to create lcd
④ solve using numerator
⑤ check for extraneous solutions!
multiple - choice question
why did we have to reject (a = -5)?
because when substituting it into the highlighted factors in the lcd, we ended up with 0 in those denominators.
because when substituting it into the highlighted factors in the lcd, we ended up with a negative value in those denominators.
rewatch

Explanation:

To determine why \( a = -5 \) is rejected, we analyze the original rational equation's denominators. The LCD is \( (a + 5)(a - 5) \). A rational expression is undefined when its denominator is zero. Substituting \( a = -5 \) into \( (a + 5) \) (a factor of the LCD) gives \( (-5 + 5)=0 \), making the original denominators (like \( a + 5 \) and \( a^2 - 25=(a + 5)(a - 5) \)) zero. So \( a = -5 \) is extraneous because it makes the equation undefined (denominators zero), not because of negative values. The first option correctly states this (substituting \( a = -5 \) into LCD factors gives zero in denominators). The second option is incorrect as it mentions a negative value, which is not the issue—undefined (zero denominator) is the reason.

Answer:

The correct option is: "Because when substituting it into the highlighted factors in the LCD, we ended up with 0 in those denominators." (the first option provided in the multiple - choice question)