Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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)

Can somebody show me how to solve this? thanks

The line intesects the x axis when y is zero. The problem is that you have to many variables.

Stmt 1 alone reduces the number of variables from 4 to 3 and is not suff.

Stmt 2 alone reduces the number of variales from 4 to 2 and is not suff.

Together you can use stmt 1 to write the equation in terms of y,x,b. You can then use stmt 2 to plug in y and x and get b. You then have the equation of the line and can find where it intersects the x axis.

y = (x + a) (x + b)
from the question, we need intercept of x, so y = 0
(x + a) (x + b) = 0
x = -a or -b
1) a+b = -1
2) lines intersects y-axis at (0,6)
so -6 =(0+a) (0+b)
ab = -6

buy solving one and two you can have x = 3, -2.
So we have two points and so I think ans is E or C???.
Pls let us know the OA.

guys, the OA to this is C. So we need to figure out how exactly is it so. would anyone please explain?

I could try

With y=(x+a)*(x+b), the intersection with X axis is obtained by using the equation of the x axis.

The X axis has y=0 for equation.

So, now, we have the following system:
o y=(x+a)*(x+b)
o y=0

That means:
(x+a)*(x+b) = 0

Before going further and keeping a concrete view of we talk about, I would like to bring your attention on y=(x+a)*(x+b). It's the equation of parabola such that y = u*x^2 + v*x + w where u = 1, making it in the shape of a valley in the XY plane.

Having said that, we have 2 cases to make (x+a)*(x+b) = 0:
> x = -a
or
> x = -b

So, we know that the 2 points are at (-a,0) and (-b,0).

All we have to do now is to look if we can have values for a and for b.

(C) it is ... We have to solve a symetrical system, from which a and b can be interverted yet we can know the 2 points (a and b are not the final things we are asked to know) :
o a+b = -1
and
o a*b = -6

guys, i understand why A and B aren't suff., however, i'm still confused with how option C is suff. can somebody show me the exact steps of the calculation in order to make option C suff?

guys, i understand why A and B aren't suff., however, i'm still confused with how option C is suff. can somebody show me the exact steps of the calculation in order to make option C suff?

(x+a)*(x+b)
= x^2 + (a+b)*x + a*b

Stat 1.... Give us a+b = -1

Stat 2... If x = 0, then y = -6....

So, y = -6 = x^2 + (a+b)*x + a*b = 0^2 + (a+b)*0 + a*b
<=> a*b = -6

Stat 1 & 2 o a + b = -1
o a*b = -6

U can solve it by replacing a or b in a*b = -6... But in the G day, u better not do so

U have 2 different equations with 2 unknown variables... and a symetrical system here from which we could not define what is a or b in certitude. Again, that's not relevant to conclude

Let me develop it anyway
o b = -1 - a = -(a+1)
o a*b = a*(-1*(a+1)) = -6
<=> -a*(a+1) = -6
<=> a^2 + a = 6
<=> a^2 + a - 6 = 0