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by resilient » Thu Apr 03, 2008 10:35 pm
if x^2 -2x-15=(x+r)(x+s) for all values of x and if r and s are constants, then which of the following is a possible value of r-s?


a.8
b.2
c.-2
d.-3
e.-5


qa is a

I dont understand the logic of the answer at all!
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by mleviko » Fri Apr 04, 2008 10:35 am
first factor the quadratic equation to:
x^2 - 2x - 15 = (x+3)(x-5)
(x+3)(x-5)=(x+r)(x+s)

then you can see that r,s equal +3,-5 or -5,+3

you then need to see what r-s can result in:
+3 - (-5) = 8
or
-5 - 3 = -8

so the only possible answer choice is (a)
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ok

by resilient » Fri Apr 04, 2008 1:23 pm
then you can see that r,s equal +3,-5 or -5,+3

I can see the r= 3 and s=-5 by putting them under eachother but

I cant see the other way around! Can you help?
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Re: ok

by Stuart@KaplanGMAT » Fri Apr 04, 2008 1:41 pm
Enginpasa1 wrote:then you can see that r,s equal +3,-5 or -5,+3

I can see the r= 3 and s=-5 by putting them under eachother but

I cant see the other way around! Can you help?
(x+r)(x+s) = (x+s)(x+r)

so, we can line up the solutions either way. We know that one of s and r is +3 and the other is -5, but we have no clue which is which. So, we could have:

s = 3, r = -5

or

r = 3, s = -5
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