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Re: DS: Coordinate Geometry (intersection) [#permalink]
19 Aug 2008, 07:58

lordw wrote:

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).

Guys I felt lost in this problem with all the letters. Any idea of how to put this in a more "visible" way? For instance, may be picking numbers? Many tks. Lw.

Graph intersects x-axis when y=0 , therefore two points where graph intersects is x=-a, x=-b

In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis? (1) a+b=-1 (2) The graph intersects y-axis at (0,-6)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis? (1) a+b=-1 (2) The graph intersects y-axis at (0,-6)

Q is in what two points graph intersect the x-axis. Which means we should be able to find two points: (x1,0) and (x2,0), so basiacally the roots of equation: (x+a)(x+b)=0.

y=(x+a)(x+b)=x^2+x(a+b)+ab. --> We must now the values of (a+b) and ab.

(1) a+b=-1 --> not sufficient. (2) x=0, y=-6 --> -6=ab not sufficient

(1)+(2) a+b=-1, ab=-6 --> y=x^2+x(a+b)+ab=x^2-x-6 --> x intersection means y=0 --> 0=x^2-x-6 --> (x+2)(x-3)=0 - -> x1=-2, x2=3. Sufficient

Answer C.

OMG!!

I didnt realize that I need to find two points!! thats why when i ended up with -2 and 3, i got confused because i was left with two values!!!

how did you find ab=-6? (2) x=0, y=-6 --> -6=ab not sufficient

Great Bunuel! I have been seeing your post and I have to say I am quite amazed at your math skills! Have you taken the GMAT yet? I would very much like to know your score.

Here, we need to find the solution for (x+a)(x+b) = 0. Solutions are x=-a, x=-b. Thus we need to find unique values for a and b to answer the question.

1) we get a relation between a and b but not sufficient to derive unique values of a and b. insuff.

2) we get a point on the graph, so substituting values of x and y in the graph we get ab = -6. Again, this is insufficient to find unique values of a and b. insuff.

Combining, we have, a+b=-1 and ab= -6, two variables, two equations. So answer is C. The points will be (-a,0), (-b,0).

In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis? (1) a+b=-1 (2) The graph intersects y-axis at (0,-6)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Answer: C.

Hope it's clear.

I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer.Can you please help on this.

In the xy-plane, at what two points the graph of y=(x+a)(x+b) intersect the x-axis? (1) a+b=-1 (2) The graph intersects y-axis at (0,-6)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.

Answer: C.

Hope it's clear.

I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer. Can you please help on this.

The question asks: at what two points the graph of \(y=(x+a)(x+b)\) intersect the x-axis?

So we should find two points: (x1, 0) and (x2, 0), (points of intersection of the given graph with X-axis). Basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\) --> \(x^2+(a+b)x+ab=0\).

When we combine statement we have: \(a+b=-1\) and \(ab=-6\), so you should solve quadratic equation \(x^2-x-6=0\) --> \(x_1=-2\) and \(x_2=3\) --> points of intersection are (-2, 0) and (3, 0). Check on the diagram:

Attachment:

MSP272919df1bed26600c1g0000673adafe5fce1fc2.gif [ 3.61 KiB | Viewed 31542 times ]

This problem is easy when you don't start by rapidly trying to solve it. Think about it before you put pen to paper... "what two points does the graph intersect the x axis..." This is a simple problem asking for 2 values of x... statement 1 gives us a+b=-1 statement 2 gives us ab

Start with the stem...multiply x+a and x+b, it will become very clear. Y = X^2 + X (a+b) + ab statement 1 gives you a+b, we need ab statement 2 says graph intersects y at (0,-6). Plug this into the original statement, you will get ab. Solved = C!

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...