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

3

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Expert's post

1

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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!

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