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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

Re: GMAT Prep Question, Any Shortcuts? [#permalink]
24 Dec 2009, 09:16

4

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

JimmyWorld wrote:

This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve?

In the xy plane, at what 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)

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.

For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).

Re: GMAT Prep Question, Any Shortcuts? [#permalink]
24 Dec 2009, 09:37

y=(x+a)(x+b) when y=0 To solve this one, what do we need to know? Obviously a or b, which are not stated in the information (1) & (2) So one rule advise by MGMAT Book, always expand when the information given is factorized or Factorized when the information given is expended. We know from a quadratic expression, the x axis intersect when y=0 So let's expand, and we have x^2 + (a + b)x + ab = 0 So now, we need to know ab and (a+b) to solve this equation Therefore the correct answer is C, since only both information taken together permit to answer the question.

Re: GMAT Prep Question, Any Shortcuts? [#permalink]
02 Nov 2010, 06:47

JimmyWorld wrote:

This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve?

In the xy plane, at what 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)

Re: Points of intersection [#permalink]
26 Oct 2011, 02:04

2

This post received KUDOS

kotela wrote:

In the XY plane 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 anyone plz explain?

Intersect the x axis means y is 0. the x coordinates will be -a and -b respectively.

1. a+b=-1 Insufficient

2. -6=(0+a)(0+b) ab=-6 Insufficient.

1+2 a+b=-1 ab=-6 a=-3, b=2 a=2, b=-3 Sufficient as we want two points we can say (-3,0) and (-2,0) without knowing the exact values of a & b. _________________

Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.

Last edited by blink005 on 26 Oct 2011, 06:17, edited 2 times in total.

Re: Intersection at X axis? [#permalink]
16 Feb 2012, 17:41

Here is my attempt:

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).

Essentially they are asking us to solve for x when y=0:

y = (x+a)(x+b) -> x^2 + xb + xa + ab = 0 (What is X?)

1) a = -b - 1 -> This is insufficient because it still leaves us with one unknown: b

2) -6 = ab -> This allows us to solve for ab in the line equation, but we still don't know what a and b are individually, which we need because we have an xb and xa term.

1) + 2) -> Knowing a = -6/b, we can plug into 1) and find b and a separately. Then the only unknown is x which we can then solve for.

Re: In the xy plane, at what points does the graph of y= [#permalink]
26 Feb 2012, 22:50

I tried plugging in values in this one.

Started from option 2: ab = -6 Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1: substitute for x and y in equation we get a+b = -1 Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.

Re: In the xy plane, at what points does the graph of y= [#permalink]
26 Feb 2012, 23:03

Expert's post

mourinhogmat1 wrote:

I tried plugging in values in this one.

Started from option 2: ab = -6 Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.

Option 1: substitute for x and y in equation we get a+b = -1 Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.

Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.

This is not a good idea to plug the numbers for this problem, it's better to understand the concept and you won't need any math at all (certainly you won't need to solve a+b=-1 and ab=-6).

Next, there are infinitely many values of a and b possible to satisfy ab=-6, not just four: notice that we are not told that a and b are integers only, so for example a=1/2 and b=-12 is also a solution. _________________

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)

So, it's important to know --- when the quadratic is given in factored form --- y= (x + a) (x + b) --- then we know the two roots, x = -a, and x = -b. Roots are the x-intercepts, the places where the graph intersects the x-axis. Basically, the prompt is asking us to find the values of a & b.

Statement #1: a + b = -1

One equation for two unknowns. Not enough to solve. Not sufficient.

Statement #2: The graph intersects the y-axis at (0, -6)

Plugging in x = 0 (the condition of the y-axis), we get y = (0+a)(0+b) = ab = -6

Again, one equation for two unknowns. Not enough to solve. Not sufficient.

Combined statements: a + b = -1 ab = -6

Two equations with two unknowns ---> we can solve for the values of a & b, which will answer the question. Sufficient.

Answer = C

Here's another practice question on quadratics for practice. http://gmat.magoosh.com/questions/120 When you submit your answer to that question, the next page will have a full video explanation.

Intersecting the x-axis meens that we have to find the roots of the equation (when will y=0?). y = (x+a) * (x+b) = x^2 + a*x + b*x + a*b = x^2 + x*(a+b) + a*b

Statement (1) is not sufficient. We don't know specific values of a and b, nor can we substitute all variables (a and b). Statement (2) just gives us one point.

In equation no 2 - How you replace ab with -6. Will you please explain the logic behind it ?

Thanks in advance.

Sure! Lets take another look at (2)The graph intersects the y axis at (0, -6 ) We know that the parabola has equation y=x^2+x(a+b)+ab, now if it intersect y in (0,-6) this point is in the parabola, so we just substitute its value (x=0,y=-6) in the equation -6=0^2+0(a+b)+ab and we find that ab=-6

Hope it's clear now, let me know _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: In the xy plane, at what points does the graph of y= [#permalink]
24 Apr 2013, 17:10

This question is a bit tricky , if you know the following it helps : Any quadratic equation represents a parabola in the x-y domain. here , the equation is y=(x+a)(x+b) . So the roots of the equation are -a,-b . So it cuts the x axis at -a and -b.

1) a+b = 1 -> Doesn't help since we have one equation with 2 variables. 2) Gives you one set of value for x and y. So you can get an equation in a and b.

Combining 1 and 2 gives the answer to A and B. C wins.

Re: In the xy plane, at what points does the graph of y= [#permalink]
25 May 2014, 03:10

I tried to get the equation by applying the special product of (x+y)^2. Is my reasoning correct that I cant do that because a is not equal to b so we basically dont have a y?!

Re: In the xy plane, at what points does the graph of y= [#permalink]
25 May 2014, 04:39

Expert's post

steilbergauf wrote:

I tried to get the equation by applying the special product of (x+y)^2. Is my reasoning correct that I cant do that because a is not equal to b so we basically dont have a y?!

Thanks in advance!

Your question is not clear at all. Can you please elaborate? Thank you. _________________

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...