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a,b,c and x

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Source: — Data Sufficiency |
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by HeintzC2 » Tue Aug 09, 2011 1:31 pm
we know that (x+a)(x+b) expands to x^2+a*x+b*x+a*b. therefore c*x=a*x+b*x or c = a+b

using option 1) a*b>0; c>a if a and b are positive, c<a if a and b are negative. insufficient.

using option 2) both a and b are less than 0. adding 2 negatives only gets more negative. a>c. sufficient.

IMO B
GmatKiss wrote:If (x + a)(x + b) = x^2 + cx + 17 and a, b, and x are integers, is a > c?

1. ab > 0
2. a < b < 0
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by ColumbiaVC » Tue Aug 09, 2011 5:13 pm
IMO D

(x+a)(x+b)=x^2+cx+17
=> x^2+x(a+b)+ab=x^2+cx+17

Comparing the coefficients on both the sides
=> ab=17 and a+b=c

Also recalling a, b, x are integers (can be +ve or -ve)

1) ab>0
(-ve)(-ve)>0 or (+ve)(+ve)>0
As a,b are integers and ab=17 the possible conditions are
=> i) a=-1 and b=-17 (-1)(-17)=17>0 then c=a+b=(-1)+(-17)=-18 Thus (-1)>(-18) a>c YES
ii) a=-17 and b=-1 (-17)(-1)=17>0 then c=a+b=(-17)+(-1)=-18 Thus (-17)>(-18) a>c YES
Sufficient

2) a<b<0
=> both a and b are -ve integers and a<b
As a,b are integers and ab=17 the only possible condition is
=> i) a=-17 and b=-1 (-17)(-1)=17>0 then c=a+b=(-17)+(-1)=-18 Thus (-17)>(-18) a>c YES
Sufficient

Hope I am not missing any fine print of the question. Correct me if I am wrong.
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by gmatboost » Tue Aug 09, 2011 10:23 pm
I agree with Heintz. To expand on the answer a bit:

We know right away that ab = 17, so either they are 1 and 17, or -1 and -17.

As Heintz shows, we can determine that a + b = c

If the numbers are 1 and 17, then a > c is false, since c = 18 and a is 1 or 17
If the numbers are -1 and -17, then a > c is true, since c = -18 and a is -1 or -17

Statement 1 [spoiler]does not help us choose between these 2 cases, since either way ab = 17 > 0[/spoiler].
Statement 2 tells us that we are talking about Case 2, so it is sufficient.
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