At what point does y=(x+a)(x+b) intersect x-axis?

[denise.lu]

DS question - In the XY plane at what point does y=(x+a)(x+b) intersect x-axis?
(1) a+b= -1
(2) graph intersects y-axis at (0, -6)

The answer is C - not too sure how to get started on this problem?

[khushbu.bhalla]

If the line intersects at x axis, which means the Y value would be 0
so (x+a)(x+b) = 0
x^2 +x(a+b) + ab = 0
from 1) a+b =-1
so x^2 - x + ab = 0 (Not sufficent)
2) Graph intersects Y axis at (0,-6) so
putting in the initial equation
we get 36 = ab
Not Sufficent

Together

we get the equation x^2 - x + 36 = 0
solving the equation we can find the value of x so , the answer is C

[shivakumars29]

Hi,,
Just wondering--how did you get ab=36 from the second case?

from the eqn- y=x^2+x(a+b)+ab
and sub for x=0 and y=-6 from statement 2, we would get ab=-6 (please help me if my understanding is wrong)

Although your final answer makes sense, as combining the two, we can get the value for x

[RonPurewal]

the last poster is correct: plugging (0, -6) into statement 2 gives ab = -6, not ab = 36.
go through the steps:
y = (x + a)(x + b)
-6 = (0 + a)(0 + b)
-6 = ab
there you go.
the 36 could just be a typographical error, but that's unlikely, as the "-" and the "3" are nowhere near each other on any keyboard i've ever seen.

also note that x^2 - x + 36 is an equation with no solutions, a fact that can easily be verified by plugging into the quadratic formula. they will NOT give a data sufficiency problem with no solution.

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the rest of the analysis is correct; the problem winds up as x^2 - x - 6 = 0, an equation that factors to (x - 3)(x + 2) = 0 and has the two solutions x = 3 and x = -2.

[fighting_cax]

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the rest of the analysis is correct; the problem winds up as x^2 - x - 6 = 0, an equation that factors to (x - 3)(x + 2) = 0 and has the two solutions x = 3 and x = -2.

I have reached up to this point in the computation. Given that the equation has 2 solutions, shouldn't the answer be E instead? The question is asking for only 1 point.

[RonPurewal]

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the rest of the analysis is correct; the problem winds up as x^2 - x - 6 = 0, an equation that factors to (x - 3)(x + 2) = 0 and has the two solutions x = 3 and x = -2.

I have reached up to this point in the computation. Given that the equation has 2 solutions, shouldn't the answer be E instead? The question is asking for only 1 point.

looks like the original poster misquoted; the question should ask for the two points at which the graph intersects the axis.

here's a reference, in which the problem is transcribed that way:
post13795.html