r and s are roots

[massi2884]

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

1) b<0
2) c<0

OA B Source OG13

[aneesh.kg]

for a quadratic equation in the form
ax^2 + bx + c = 0,
Product of roots = c/a

If r and s are the roots of
x^2 + bx + c = 0, then
rs = c/1 = c

Statement(1):
rs is not dependent on b.
INSUFFICIENT

Statement(2):
c < 0, so rs < 0
SUFFICIENT

[spoiler](B)[/spoiler] is the answer

[neelgandham]

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

If r and s are roots of the equation, then the equation is of the form (x-r)*(x-s)=0
i.e ((x-r)*x)-((x-r)*s)=0
x^2 -r*x -s*x + rs = 0
x^2 -(r+s)*x + rs = 0 = x^2 + bx + c (From the question)
Equation the coefficients of x and constants
-(r+s) = b and rs = c

1) b<0

-(r+s)<0
(r+s)>0
If r = 1 and s = 2((r+s)>0), rs = 2, which is greater than 0
If r = -1 and s = 2((r+s)>0), rs = -2, which is less than 0.
So, statement I is insufficient to answer the question.

2) c<0

c = rs < 0
So, statement II is sufficient to answer the question.

IMO B

[Brent@GMATPrepNow]

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

1) b<0
2) c<0

OA B Source OG13

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