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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
24 Dec 2009, 09:16
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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: Line intersection x-axis [#permalink]
03 May 2011, 17:33
Expert's post
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udaymathapati wrote:
Attachment:
M-Q19.JPG
At what two points does the graph of y = (x+a)(x+b) intersect the x axis?
You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.
0 = (x+a)(x+b) or x = -a or -b Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.
Statement 1: a + b = -1 Two variables, only one equation. Not sufficient.
Statement 2: Graph intersects the y axis at (0, -6). At y axis, x = 0. This means when x = 0, y co-ordinate is -6. Put these values in y = (x+a)(x+b) to get -6 = ab. Again, two variables, one equation. Not sufficient alone.
Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.
Re: Points of intersection [#permalink]
26 Oct 2011, 02:04
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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
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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.
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