Quadratic equation - DS question

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Quadratic equation - DS question

by GMAT4321 » Thu Nov 26, 2015 11:05 am
Hello,

Could someone please shed some light on how to solve the below question. I suppose it could be solved with a lot of trial and error, but I'm wondering if I'm missing something that can help me solve it within 2-3 mins. Is there a general rule about quadratics that could help?

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

Thanks for the help!

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by Brent@GMATPrepNow » Thu Nov 26, 2015 11:08 am
If r and s are the roots of the equation x² + bx + c = 0, where b and c are constants, is rs < 0?

1) b < 0
2) c < 0
Let's first examine the relationship between the roots of an equation and the given equation. Here are some examples:

Example #1: x² - 5x + 6 = 0
We can rewrite this as x² + (-5x) + 6 = 0 [to make it look like the given equation x² + bx + c = 0]
So, b = -5 and c = 6
To solve the equation, we'll factor to get: (x - 3)(x - 2) = 0
So, the ROOTS of the equation are x = 2 and x = 3
NOTICE that the sum of the roots equals -b, and notice that the product of the roots = c

Example #2: x² + 6x - 7 = 0
We can rewrite this as x² + 6x + (-7) = 0 [to make it look like the given equation x² + bx + c = 0]
So, b = 6 and c = -7
To solve the equation, we'll factor to get: (x + 7)(x - 1) = 0
So, the ROOTS of the equation are x = -7 and x = 1
NOTICE that the sum of the roots equals -b, and notice that the product of the roots = c

We could keep going with more examples, but the big takeaway is as follows:
If r and s are the roots of the equation x² + bx + c = 0, then r + s = -b, and rs = c

Okay, now onto the question....

Target question: Is rs < 0?

Given: r and s are the roots of the equation x² + bx + c = 0

Statement 1: b < 0
This means that b is NEGATIVE, which also means that -b is POSITIVE
From our conclusions above, we saw that r + s = -b
So, we can now conclude that r + s = some POSITIVE VALUE.
Is this enough info to determine whether rs < 0?
NO.
Consider these two conflicting cases:
Case a: r = -1 and s = 2 (here r + s = some positive value), in which case rs < 0
Case b: r = 1 and s = 2 (here r + s = some positive value), in which case rs > 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: c < 0
From our conclusions above, we saw that rs = c
Now, statement 2 tells us that c is negative.
So, it MUST be the case that rs < 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMAT4321 » Sat Nov 28, 2015 5:23 am
Thanks very much for your comprehensive response Brent :)