massi2884 wrote: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
