Data Sufficiency

1. If r and s are the roots of the equation X^2+bx+c=0, where b and c are constants, is rs<0?
1) b<0
2) c<0

I don't understand how the correct answer is b when looking at the answer key - looking for a clearer explanation.


2. The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?

1) x>y
2) xy/100<x-y

Not sure how to approach this in the easiest way. Explanation seemed very long and I don't fully understand it. Can you please provide your expertise on how to handle this question?


Also, question #136 DS - from the 13th Edition GMAT review - I don't get. I can't type the symbols, so if I can get clarification on how to attack that question, I would really appreciate it.

Thank you GMAT experts!

B

If r and s are the roots of the equation x^2+bx+c=0, where b and c are constants, is rs<0?
1) b<0
2) c<0

In order for rs to be < 0, it must be the case that one of the variables is positive and the other negative. So the question is really asking "Do r and s have opposite charges?"

Think about some simple quadratic equations when that last term is negative:

x^2 - 2x - 15 = 0
(x - 5)(x+3) = 0
So our roots are x = 5 or x = -3.

x^2 +4x - 21 = 0
(x + 7)(x-3) = 0
So our roots are x = -7 or x =3.
Again, we have one positive and one negative root.

Notice that if the last term is negative, we'll always have one positive and one negative root. (When we're factoring, we're trying to come up with two numbers that will multiply to that last term. Only a positive and negative will give a negative product.)


Let's see what happens when that last term is positive:

x^2 - 6x + 8 = 0
(x - 4)(x-2) = 0
x = 4 or x = 2.
Two positive roots.

Or:

x^2 + 6x + 8 = 0
(x+4)(x+2) = 0
x = -4 or x = -2
Two negative roots.

Notice that if the last term of the quadratic is positive, we'll end up with either two positive roots or two negative roots. Again, this makes sense. We want two numbers that will multiply to give us a positive product and either two positives or two negatives will work.

S1: b < 0.
So the coefficient of the middle term is negative.

Equation 1: x^2 - 2x - 15 = 0.
(x - 5)(x+3) = 0, so x = 5 or x = -3.
[Because r and s are defined as the roots, the values for x, which are the roots of the quadratic, are giving us our values for r and s]
We get one positive and one negative so rs < 0, and the answer is YES

Equation 2: x^2 - 6x + 8 = 0.
(x - 4)(x-2) = 0, so x = 4 or x = 2.
Both r and s are positive so rs > 0 and the answer is NO.

Not sufficient.

S2: Now we know that the last term is negative. We also know that, when factoring the quadratic, we need two terms to multiply to that last term. The only way this can happen is if we have a positive and negative.

Maybe you'd run through a few examples to prove this to yourself:

x^2 - 2x - 15 = 0
(x - 5)(x+3) = 0
So our roots are x = 5 or x = -3.
One positive, one negative root, so rs < 0, and we have a YES



x^2 +4x - 21 = 0
(x + 7)(x-3) = 0
So our roots are x = -7 or x =3.
One positive, one negative root, so rs < 0, and we have a YES

So long as that last term is negative, we'll always have roots with opposite signs, so this statement is sufficient.

The answer is B.