Target question: Is rs < 0?Val911 wrote: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
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
