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Algebra

by [email protected] » Sat Apr 11, 2015 12:48 pm
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.


Thank you GMAT experts!

B
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by Brent@GMATPrepNow » Sat Apr 11, 2015 1:24 pm
[email protected] wrote: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
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by GMATGuruNY » Sat Apr 11, 2015 1:43 pm
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
Approach 1:

For any quadratic in the form ax² + bx + c = 0:
The sum of the roots = -b/a.
The product of the roots = c/a.

Here, a=1, so the product of the roots = c/1 = c.

Question rephrased: Is c<0?

Statement 1: b<0
No information about c.
INSUFFICIENT.

Statement 2: c<0
SUFFICIENT.

The correct answer is B.

Approach 2:

Since the question stem asks about the product of the roots, and the two statements offer information about the signs of b and c, test four cases:

Case 1: (x+2)(x+1) = x² + 3x + 2.
Here, rs = (-2)(-1) = 2, b=3, c=2.

Case 2: (x+2)(x-1) = x² + x - 2.
Here, rs = (-2)(1) = -2, b=1, c=-2.

Case 3: (x-2)(x+1) = x² - x - 2.
Here, rs = (2)(-1) = -2, b=-1, c=-2.

Case 4: (x-2)(x-1) = x² - 3x + 2.
Here, rs = (2)(1) = 2, b=-3, c=2.

Statement 1: b<0
b<0 in Case 3 and Case 4.
In Case 3, rs<0.
In Case 4, rs>0.
INSUFFICIENT.

Statement 2: c<0
c<0 in Case 2 and Case 3.
In both cases, rs<0.
SUFFICIENT.

The correct answer is B.
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by ceilidh.erickson » Mon Apr 20, 2015 11:44 am
Another approach, building on Brent's explanation, is this:

If the roots of the equation are r and s, we could work backwards to say:
(x - r)(x - s) = 0

Then FOIL this out:
x² - sx - rx + rs = 0

Factor out the x to get it into the format x² + bx + c = 0:
x² + (-r - s)x + rs = 0

It's clear that b = (-r - s) and c = rs. Thus, we can rephrase the question is rs < 0? as is c < 0?

Clearly, statement 2 is sufficient but statement 1 is not.
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