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SajjadAhmad
If \(\frac{a^2 -9}{12a} =\) \(\frac{a - 3}{a + 3} , a + 3\) # 0, and a # 0, then \(a =\)?

(A) 1
(B) 2
(C) 3
(D) 6
(E) 9

Source: Nova GMAT

simplify for given expression \(\frac{a^2 -9}{12a} =\) [m]\frac{a - 3}{a + 3}
we get a^2-6a+9=0
a= 3
IMO C
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SajjadAhmad
If \(\frac{a^2 -9}{12a} =\) \(\frac{a - 3}{a + 3} , a + 3\) # 0, and a # 0, then \(a =\)?

(A) 1
(B) 2
(C) 3
(D) 6
(E) 9

Source: Nova GMAT

Cross multiplying, we have:

(a + 3)(a - 3)(a + 3) = 12a(a - 3)

(a - 3)(a + 3)^2 - 12a(a - 3) = 0

(a - 3)[(a + 3)^2 - 12a] = 0

At this point, we see that if a - 3 = 0, or a = 3, the equation will be true.

Answer: C
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