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In the xy plane, at what points does the graph of y= [#permalink]
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In the xy plane, at what points does the graph of y=(x+a)(x+b) intersect the xaxis? (1) a + b = 1 (2) The graph intersects the y axis at (0,6)
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Originally posted by JimmyWorld on 24 Dec 2009, 10:00.
Last edited by Bunuel on 26 Feb 2013, 05:23, edited 2 times in total.
Edited the question and added the OA



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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
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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 xaxis? (1) a + b = 1 (2) the graph intersects the y axis at (0,6) OA Thanks. Xintercepts 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 \((x+a)(x+b)=0\). \((x+a)(x+b)=0\) > \(x^2+bx+ax+ab=0\) > \(x^2+(a+b)x+ab=0\). 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 yintercept, 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 xintercepts of the given graph. Answer: C. For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature). Hope it helps.
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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
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24 Dec 2009, 10:37
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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.



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Re: DS question : need help [#permalink]
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15 Oct 2010, 01:05
satishreddy wrote: need help on DS question Right, so when the graph intersects the xaxis, y is 0. Those points are clearly x=a and x=b (1) a+b=1. Not sufficient to know a or b (2) We have point (0,6). Plus it into the equation to get 6=ab. Again not sufficient to know either number (1+2). Now we have 2 equations, sufficient. a6/a=1; a^2a6=0; a=3 or a=2 Hence b=2 or b=3 .. Either way the points are 3 and 2 Answer is (c)
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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
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02 Nov 2010, 07:47
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 xaxis? (1) a + b = 1 (2) the graph intersects the y axis at (0,6) OA Thanks. We know that the graph will intersect the xaxis when y=0, which occurs when x = a and x = b. 1) This doesn't tell us what a and b are. For example, a could be 5 and b could be 6, or a could be 4 and b could be 3. Insufficient. 2) This tells us that either a or b is 6, but we don't know the other point. Insufficient. Together, we know that one of the values is 6, and that a + b = 1. So we can get the other value. Sufficient.



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Re: Line intersection xaxis [#permalink]
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03 May 2011, 10:19
y = x^2 + (a+b)x + ab is a parabola.Intersecting x axis at two points. hence we need to find the roots of this equation. a. a+b = 1 means a= 1  b = 1  a=b= 0.5 each etc. Hence not sufficient. b y = 6 for x = 0 means ab = 6 (substituting in the original quadratic equation. a= 3,b=1  a = 6, b = 1 etc. hence not sufficient. together 1+2 y = x^2 x 6 = (x3) * (x + 2) hence two points for X axis. thus C.
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Re: Line intersection xaxis [#permalink]
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03 May 2011, 10:21
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y=(x+a)(x+b) = x^2+(a+b)x+ab this is parabola equation. intersects xaxis means y=0
st 1> a+b = 1 x^2x+ab=0 cannot solve the equation; not sufficient
st 2> graph intersects yaxis at 6 i.e. ab = 6 x^2+(a+b)6=0 cannot solve the eq, not sufficient.
Both together, x^2x6=0 Solving gives x = 2 or x=3
Hence C.



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Re: Line intersection xaxis [#permalink]
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03 May 2011, 18:33
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udaymathapati wrote: 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 coordinate 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 coordinate 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. Answer (C).
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Re: coordinate geom [#permalink]
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25 May 2011, 03:26
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To find xintercepts, you set y=0 and solve. The question asks for the xintercepts of y = (x+a)(x+b). Setting y=0, the question is asking when (x+a)(x+b) = 0. Since one of the factors on the left side must be 0, we find we have intercepts when x = a and x = b. Statement 1 tells us the sum of a and b, which doesn't let us find their values alone. Statement 2 gives us a point on the curve; if we plug that point into our equation, what we get must be true. So 6 = (0+a)(0+b), and ab = 6. So Statement 2 gives us the product of a and b, which isn't enough to find their individual values. Together, we know that a+b = 1 and ab = 6. That is, we know the sum and product of two numbers. Notice this is what happens every time you factor a quadratic: you look for two numbers with a certain product and a certain sum. For example, if you were to factor the quadratic x^2  x  6, you'd look for two numbers which multiply to 6 and add to 1. So if a+b = 1 and ab = 6, then our solutions for a and b must be 3 and 2, in some order. Since our intercepts are at a and b, the intercepts are at x=3 and x=2, so the information is sufficient. Notice we can't figure out which value is equal to a and which is equal to b, but we don't need to do that to answer the question; we just need to know where the two points are, and not which point is a and which point is b.
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Re: geometry [#permalink]
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24 Jul 2011, 16:37
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tt2011 wrote: In xy plane, at what 2 points does the graph of y=(x+a)(x+b) intersect the xaxis?
(1) a+b= 1
(2) The graph intersects the yaxis at (0,6)
Correct Answer: C
Please provide reasoning..thanks.. Please don't reveal your answer in the question. Hide it with spoiler. thanks 1) y=(x+a)(x+b) Question is asking for the solution of x (x+a)(x+b)=0 x^2+ax+bx+ab=0 x^2+(a+b)x+ab=0 1) a+b=1 We got a+b. Yet, we don't know about "ab" Not Sufficient. 2) Intersect y at (0,6) x=0; y=6 6=(0+a)(0+b) ab=6 We got ab. We don't know a+b. Together; Sufficient. Ans: "C"
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Re: geometry [#permalink]
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24 Jul 2011, 21:22
We need to find the values of x when x = a or when x =b
Stmt 1: a+b = 1 gives numerous values of a and b . Insuff
Stmt 2: plugging in (0, 6) in the equation:
ab= 6 still cannot determine values of x
Combining 1 and 2 gives values of x as a quadratic eqn forms:
x^2x6=0. Hence C



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Re: Points of intersection [#permalink]
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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 yaxis 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.
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Originally posted by blink005 on 26 Oct 2011, 03:04.
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Re: Intersection at X axis? [#permalink]
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16 Feb 2012, 18:41
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.
Answer: C



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Re: In the xy plane, at what points does the graph of y= [#permalink]
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26 Feb 2012, 23:50
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.



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Re: In the xy plane, at what points does the graph of y= [#permalink]
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27 Feb 2012, 00:03
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.
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Re: In the xy plane, at what points does the graph of y= [#permalink]
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26 Apr 2012, 21:33
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y= x^2+ax+bx+ab y= x^2+x(a+b)+ab
the question asks us to find the value of two points where the graph intersects x axis. So y=0
x^2+x(a+b)+ab=0 =?
1) a+b=1 , but we dot have any value of ab, hence not sufficient
2) putting the value x=0 , y=6 y= x^2+x(a+b)+ab ab=6 but we dot have any value of a+b , hence not sufficient
1&2) x^2+x(a+b)+ab=0 now we have a+b=1 ab =6 hence sufficient to find 2 values of x. So C is the answer



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Re: Question [#permalink]
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30 Apr 2012, 12:17
Stiv wrote: In the xyplane, 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 yaxis 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 xintercepts, the places where the graph intersects the xaxis. 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 yaxis at (0, 6) Plugging in x = 0 (the condition of the yaxis), 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 = CHere's another practice question on quadratics for practice. http://gmat.magoosh.com/questions/120When you submit your answer to that question, the next page will have a full video explanation. Let me know if you have any further questions. Mike
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Re: Coordinate DS; y = (x+a)(x+b) [#permalink]
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02 Sep 2012, 11:50
Intersecting the xaxis 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.
Combining both statements: 0 = (6)^2 + (6)*(1) + a*b 0 = 36 + 6 + a*b 42 = a*b
Substituting a*b (and (a+b)): x^2 + x*(1) + (42) = 0
You don't have to calculate any further. It's easy to see now that we have sufficient information to calculate the other point.



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Re: In the xy plane, at what points does the graph of y= [#permalink]
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04 Sep 2012, 01:38
1 statement is clearly insufficient, since we have nothing about x or y 2 statement tells us that the ab=6, but nothing about a or b individually Combining statements we have: a=6/b, 6/b+b=1, b^2+b6=0, a=3, b=2, so both statements together are sufficient (C)
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Re: Coordinate geometry [#permalink]
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23 Apr 2013, 01:44
kabilank87 wrote: In the xy plane at what 2 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 ) The function is a parabola \(y=x^2+x(a+b)+ab\) 1)Does \(y=x^2x+ab\) intersec x? It depend on ab.Not sufficient 2) Tells us that point (0,6) is on the parabol, so \(ab=6\). Does \(y=x^2+x(a+b)6\) intersect x? It depends on a+b. Not sufficient 1+2) The equation of the parabola is complete \(y=x^2x6\), we can tell if it intersect x. Sufficient C
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