Question

lesson 2.5 review - solving rational equations
2
\\(\frac{2a^2}{a^2 - 25} - \frac{a}{(a + 5)} = \frac{7}{(a - 5)}\\)
\\(\frac{2a^2}{(a + 5)(a - 5)}\\)
① factor denominators
② find lcd
③ multiply to create lcd
④ solve using numerator
⑤ check for extraneous solutions!
18 multiple - choice question
the reason why the factors of \\(a^2 - 25\\) are exactly \\((a + 5)(a - 5)\\) is because
-5 and 5 multiply to -25 and add up to 1 which, the middle term in this case is 1a.
-5 and 5 multiply to -25 and add up to 0 which, the middle term in this case is 0a.
-5 and 5 add up to -25 and multiply to 0 which, the middle term in this case is 0a.
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Explanation:

To determine why \(a^2 - 25\) factors to \((a + 5)(a - 5)\), we analyze the quadratic \(a^2 - 25\). This is a difference of squares, where \(a^2 - 25=a^2 - 5^2\). The formula for factoring a difference of squares is \(x^2 - y^2=(x + y)(x - y)\). Here, \(x = a\) and \(y = 5\), so \(a^2 - 5^2=(a + 5)(a - 5)\). Also, when we expand \((a + 5)(a - 5)\) using the FOIL method: \((a)\times(a)+(a)\times(- 5)+5\times(a)+5\times(-5)=a^2-5a + 5a-25=a^2-25\). The middle terms \(-5a\) and \(+5a\) add up to \(0\), so the middle term in the quadratic \(a^2 - 25\) is \(0a\) (or just not present as a linear term). Now let's analyze the options:

• First option: Says the middle term is \(1a\), which is incorrect as we saw the middle terms cancel out.
• Second option: Says \(-5\) and \(5\) multiply to \(-25\) (since \((-5)\times5=-25\)) and add up to \(0\) (since \(-5 + 5 = 0\)), and the middle term is \(0a\), which matches our analysis.
• Third option: Says \(-5\) and \(5\) add up to \(-25\), which is incorrect (they add up to \(0\), and multiply to \(-25\)).

Answer:

The option (assuming the second option in the list) which states: "-5 and 5 multiply to -25 and add up to 0 which, the middle term in this case is 0a."