Problem from mba.com practice test 1

ksutthi

Can anyone help me solve this question? I don't seem to be able to derive the answer.

In the xy-plane, at what two points does the graph of y = (x+a)(x+b) intersect the x-axis?

(1) a+b = -1
(2) The graph intersects the y-axis at (0,-6)

Many thanks.

ed09 · Posted: Sat Jul 05, 2008 9:28 pm Post subject:

I dare to say that the answer is C.

Below is my solution.

Rewrite the equation in form of quadratic
y=x^2+(a+b)x+ab

0=x^2+(a+b)x+ab - real solutions of this equation are x-coordinates of points where the graph intersects the x-axis.

(1)
a+b=-1 => x^2-x+ab=0
a+b=-1 - does not give enough data to be sure about ab
INSUFFICIENT

(2) (0,6) - this is one root
0=x^2+(a+b)x+ab; (x-6)(x+z)=0; x2+zx-6x-6z=0; x2+(z-6)x-6z=0
there is no enough data to solve this equation further
INSUFFICIENT

(3) COMBINE
from (2) x2+(z-6)x-6z=0
from (1) 0=x^2+(a+b)x+ab; a+b=-1
=>(a+b)=z-6=-1 => z=5 => -6z=-30
hence the equation sought is x2-x-30=0 or (x-6)(x+5)=0
thus, roots of (0,6) and (0,-5) are coordinates of points where the graph intersects the x-axis.
SUFFICIENT

I'll appreciate someone, who find a mistake in my solution or show a shorter way.
Best!

ksutthi · Posted: Sat Jul 05, 2008 10:44 pm Post subject:

ed09 - yes, you got it. the answer is C.

Many thanks!

Carol · Posted: Wed Jul 09, 2008 12:15 pm Post subject:

ed09

since you got it, could you try to explain your reasoning, elaborate it bit longer and clearly.

I didn't quite understood your explanation regarding the 2nd sentence.

Many thanks

ksh · Posted: Thu Jul 10, 2008 3:39 am Post subject:

ed09 · Posted: Fri Jul 11, 2008 1:05 pm Post subject:

Hi ksh, or Rising GMAT Star, Wink

I truly appreciate your remark, since it correctly points to my mistake.
I was definitely wrong in solving the second statement.

Indeed, the statement "The graph intersects the y-axis at (0,-6)"
means that x=0, but y=-6. Therefore, after substitution of these variables into the initial equation we get -6=0^2+(a+b)0+ab
That means -6=ab.
This also doesn't let us to find the second point of the graph's intersecting the x-axis.
That is the second statement is still INSUFFICIENT.

However, when we combine the statements and substitute their results
a+b=-1
ab=-6
in the original equation we get
0=x^2+(-1)x+(-6)
That is x^2-x-6=0 or (x+2)(x-3)=0. As a result of solving this we can find two roots: x=-2 or x=3.
Hence, the points of (-2,0) and (3,0) are those which we are asked about.
SUFFICIENT

The answer is still C.

Sorry for my pointing to that, but I just think it's worth to notice that your last equation and the remark about it are not accurate.
The equation of x^2-x+6=0 has no real solutions, and so we can't apply them to the xy-plane given.
I believe, you've realized this already.

Anyway, thanks a lot, and good luck on the test!