how is the answer C? can someone please explain?

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.

The answer should be -16. From the first root we know that the value of a is -8. Now they say that the above equation has equal roots so -15 fails as for 2nd equation the roots were 3 and 5,with 16 we get roots 4,4

The equation \(x^2 + ax - b = 0\) has equal roots, and one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3. What is the value of b?

A. \(-\frac{1}{64}\)
B. \(-\frac{1}{16}\)
C. \(-15\)
D. \(-16\)
E. \(-64\)


Since one of the roots of the equation \(x^2 + ax + 15 = 0\) is 3, then substituting we'll get: \(3^2+3a+15=0\). Solving for \(a\) gives \(a=-8\).

Substitute \(a=-8\) in the first equation: \(x^2-8x-b=0\).

Now, we know that it has equal roots thus its discriminant must equal to zero: \(d=(-8)^2+4b=0\). Solving for \(b\) gives \(b=-16\).


D is the answer

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.