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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]

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24 Dec 2009, 10:37

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]

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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 x-axis?

(1) a + b = -1 (2) the graph intersects the y axis at (0,-6)

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.

To find x-intercepts, you set y=0 and solve. The question asks for the x-intercepts 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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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.
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Last edited by blink005 on 26 Oct 2011, 07:17, edited 2 times in total.

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]

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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.

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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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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