Yes/No DS

[Val911]

Hey Guys,

Stumbled upon the following question:

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

1) b<0

2) c <0

Any takers?

[Brent@GMATPrepNow]

Hey Guys,

Stumbled upon the following question:

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

Target question: Is rs < 0?

IMPORTANT: If r and s are the roots of the equation x^2 + bx + c = 0, then we can rewrite x^2 + bx + c as (x - r)(x - s)

Example: If 5 and -2 are the roots of the equation x^2 - 3x - 10 = 0, then it must be true that x^2 - 3x - 10 = (x - 5)(x + 2)

So, we now know that x^2 + bx + c = (x - r)(x - s)
Expand the right side to get: x^2 + bx + c = x^2 - sx - rx + rs
Simplify right side to get: x^2 + bx + c = x^2 + (-s-r)x + rs

So, we can see that b = -s-r, and c = rs

Statement 1: b < 0
Since b = -s-r, we now now that -s-r < 0
However, this tells us nothing about whether or not rs is less than 0
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: c < 0
Since c = rs, we can now see that rs < 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent

[GMATGuruNY]

Hey Guys,

Stumbled upon the following question:

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

1) b<0

2) c <0

Any takers?

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.

[GMATGuruNY]

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

1) b<0

2) c <0

Below is an alternate approach.

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.