Re: GRE Math Challenge #52- equation x^2 + ax - b = 0 has
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03 Apr 2019, 03:31
I believe the answer should be B.
First, we know that 3 is a root of the second equation. So when x = 3, the equation should be valid. Hence:
\(3^{2}+a3 +15= 0\)
\(3a + 24 = 0\)
\(3a = -24\)
\(a = -8\)
Now we plug that into our second equation:
\(x^{2}-8x - b = 0\)
Since we know this equation has equal roots, it must be true that (x - r)(x - r) = 0 for some r must be true. Since our middle term is -8, r must equal 4. Hence, our factorization is:
\((x - 4)(x - 4) = x^{2}- 8x + 16 = 0\)
We see that our final term here is 16. This means b = -16.