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?
Yes/No DS
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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
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For any quadratic in the form ax² + bx + c = 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 constancts, is rs < 0?
1) b<0
2) c <0
Any takers?
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.
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Below is an alternate approach.Val911 wrote::
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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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