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by shibal » Mon Jul 13, 2009 6:54 pm
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Source: — Data Sufficiency |
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by tryin700 » Mon Jul 13, 2009 9:35 pm
Given equation can also be written as

y=x^2+x(a+b)+ab --eq1

1) a+b=-1

hence y=x^2-x+ab

when y=0
x^2 - x = -ab --eq2
we dont know ab or a or bhence INSUFF

2) substituting x=0 and y=-6 in eq1

-6 = 0+0+ab

Still INSUFF

combining we get

from eq1 and eq2
x^2 - x = 6

or x^2-x-6=0

we can solve for x hence SUFF

hence C
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by ash_maverick » Mon Jul 13, 2009 11:49 pm
Hey, can you please tell me , why did u try with putting Y=0 for statement 1 in line 3. I am not able to get the reason for this substitution.

tryin700 wrote:Given equation can also be written as

y=x^2+x(a+b)+ab --eq1

1) a+b=-1

hence y=x^2-x+ab

when y=0
x^2 - x = -ab --eq2
we dont know ab or a or bhence INSUFF

2) substituting x=0 and y=-6 in eq1

-6 = 0+0+ab

Still INSUFF

combining we get

from eq1 and eq2
x^2 - x = 6

or x^2-x-6=0

we can solve for x hence SUFF

hence C
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by sreak1089 » Tue Jul 14, 2009 4:00 am
When the graph of y = (x+a)(x+b) meets X-axis, y value will be equal to ZERO. Hence y was equated to ZERO.

PS: y value will always be ZERO on X-axis.
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