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Gmat Prep Coordinate Geometry DS

by tonebeeze » Sun Jan 02, 2011 6:41 pm
I don't understand how to solve this problem. Please advise. Thanks!

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)
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by anshumishra » Sun Jan 02, 2011 6:52 pm
tonebeeze wrote:I don't understand how to solve this problem. Please advise. Thanks!

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)
y = (x+a)(x+b) = x^2 + x(a+b) + ab

a = ? , b =?

Statement 1:

a+b = -1 ----- 1
Insufficient, can't find a and b based on this fact alone.

Statement 2:
0 = 36 + 6(a+b) + ab ----2 [MADE A MISTAKE HERE..... should be 6=0+0+ab]
Insufficient.

Combined 1 and 2 :
Using equation 1 in 2 =>

36-6 + ab = 0 => ab = 30 --- 3

Since equation 1 and equation 3 are two independent equations in 2 variables [ANOTHER MISTAKE, Make sure a and b has unique value...ab is in 2nd degree so will have 2 solutions.], Hence they are SUFFICIENT.

C
Last edited by anshumishra on Mon Jan 03, 2011 3:03 pm, edited 1 time in total.
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by Whitney Garner » Mon Jan 03, 2011 8:20 am
tonebeeze wrote:I don't understand how to solve this problem. Please advise. Thanks!

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)
** Quick note - the original problem actually puts the y-intercept at (0,-6), otherwise we cannot solve.
anshumishra wrote: Statement 2:
0 = 36 + 6(a+b) + ab ----2
Be careful here - the statement tells us that the Y-intercept is (0,-6), this means that when X=0, Y=-6, the solution provided gives the equation at the point (-6,0).

The correct equation is now:
-6 = (0+a)(0+b) --> -6=ab. Also INSUFFICIENT.

Solving your first equation (a+b=-1) for a, we find that a=-1-b. Combining this with -6=ab, we get that -6=(-1-b)*b, expand that bring everything to one side we find b^2 + b - 6 = 0 --> (b+3)(b-2)=0, b=-3 or =2. Because we could have solved the first equation for b and then solved to get a=-3 or =2, it is irrelevant which variable we assign -3 and which we assign 2 --> the x-intercepts will be at -3 and 2. We can also see this when we plug the b values back into the first equation.

When b=-3, a=-1-(-3) = 2 ---> intercepts are -3, 2
When b=2, a=-1-(2) = -3 ---> intercepts are -3, 2
SUFFICIENT
anshumishra wrote:Since equation 1 and equation 3 are two independent equations in 2 variables, Hence they are SUFFICIENT.
C
Be very careful again here. Your equation 1 (a+b=-1) and Equation 3 (ab=30) are 2 independent equations in 2 variables but they are not both LINEAR equations, therefore we cannot be guaranteed to find exact values for those two variables. In fact, if you notice from the solution above, we actually do not know what a or b equals specifically, but thankfully it does not matter in the context of this problem. Imagine we had a problem where a had been the width of a container and b had been the length. If the question had asked for only the width, we would not have been able to determine if it were -3 or 2, so even together the statements would have been INSUFFICIENT.

Just remember that the rule for 2 independent equations applies only to linear equations (no multiplication of the 2 variables, no exponents on the variables, and no variables in denominators).

Hope this clears that up a bit!
:)
Whit
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by anshumishra » Mon Jan 03, 2011 10:23 am
Whitney Garner wrote:
tonebeeze wrote:I don't understand how to solve this problem. Please advise. Thanks!

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)
** Quick note - the original problem actually puts the y-intercept at (0,-6), otherwise we cannot solve.
anshumishra wrote: Statement 2:
0 = 36 + 6(a+b) + ab ----2
Be careful here - the statement tells us that the Y-intercept is (0,-6), this means that when X=0, Y=-6, the solution provided gives the equation at the point (-6,0).

The correct equation is now:
-6 = (0+a)(0+b) --> -6=ab. Also INSUFFICIENT.

Solving your first equation (a+b=-1) for a, we find that a=-1-b. Combining this with -6=ab, we get that -6=(-1-b)*b, expand that bring everything to one side we find b^2 + b - 6 = 0 --> (b+3)(b-2)=0, b=-3 or =2. Because we could have solved the first equation for b and then solved to get a=-3 or =2, it is irrelevant which variable we assign -3 and which we assign 2 --> the x-intercepts will be at -3 and 2. We can also see this when we plug the b values back into the first equation.

When b=-3, a=-1-(-3) = 2 ---> intercepts are -3, 2
When b=2, a=-1-(2) = -3 ---> intercepts are -3, 2
SUFFICIENT
anshumishra wrote:Since equation 1 and equation 3 are two independent equations in 2 variables, Hence they are SUFFICIENT.
C
Be very careful again here. Your equation 1 (a+b=-1) and Equation 3 (ab=30) are 2 independent equations in 2 variables but they are not both LINEAR equations, therefore we cannot be guaranteed to find exact values for those two variables. In fact, if you notice from the solution above, we actually do not know what a or b equals specifically, but thankfully it does not matter in the context of this problem. Imagine we had a problem where a had been the width of a container and b had been the length. If the question had asked for only the width, we would not have been able to determine if it were -3 or 2, so even together the statements would have been INSUFFICIENT.

Just remember that the rule for 2 independent equations applies only to linear equations (no multiplication of the 2 variables, no exponents on the variables, and no variables in denominators).

Hope this clears that up a bit!
:)
Whit
Thanks Whitney !
Although, I can't predict the right problem (that the intercept was to be -6 instead of 6 :) ), I definitely made a mistake when I didn't bother to check values of a and b. For sure, ab=k is a equation in 2-degree so, it is most likely to have a two values for a and b (which luckily converged to a single value here) . Conceptually, I was quite aware of it.

Thanks for pointing that out ! Hopefully, I don't miss it the next time.
Thanks
Anshu

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by Whitney Garner » Mon Jan 03, 2011 10:36 am
anshumishra wrote: Thanks Whitney !
Although, I can't predict the right problem (that the intercept was to be -6 instead of 6 :) ), I definitely made a mistake when I didn't bother to check values of a and b. For sure, ab=k is a equation in 2-degree so, it is most likely to have a two values for a and b (which luckily converged to a single value here) . Conceptually, I was quite aware of it.

Thanks for pointing that out ! Hopefully, I don't miss it the next time.
and we definitely want to be careful that we don't mix up Y-intercepts with X-intercepts! It can be an easy mistake to make so we have to stay on our toes :)

Whit
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by anshumishra » Mon Jan 03, 2011 10:59 am
Whitney Garner wrote:
anshumishra wrote: Thanks Whitney !
Although, I can't predict the right problem (that the intercept was to be -6 instead of 6 :) ), I definitely made a mistake when I didn't bother to check values of a and b. For sure, ab=k is a equation in 2-degree so, it is most likely to have a two values for a and b (which luckily converged to a single value here) . Conceptually, I was quite aware of it.

Thanks for pointing that out ! Hopefully, I don't miss it the next time.
and we definitely want to be careful that we don't mix up Y-intercepts with X-intercepts! It can be an easy mistake to make so we have to stay on our toes :)

Whit
Totally agree ! Thanks !!!
Thanks
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by tonebeeze » Mon Jan 03, 2011 7:13 pm
Whitney Garner wrote:
anshumishra wrote: Thanks Whitney !
Although, I can't predict the right problem (that the intercept was to be -6 instead of 6 :) ), I definitely made a mistake when I didn't bother to check values of a and b. For sure, ab=k is a equation in 2-degree so, it is most likely to have a two values for a and b (which luckily converged to a single value here) . Conceptually, I was quite aware of it.

Thanks for pointing that out ! Hopefully, I don't miss it the next time.
and we definitely want to be careful that we don't mix up Y-intercepts with X-intercepts! It can be an easy mistake to make so we have to stay on our toes :)

Whit
I still don't fully understand problem. maybe it would help if someone would explain the problem to me without the abbreviations. I would really like to get this problem nailed down. thanks
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by anshumishra » Mon Jan 03, 2011 7:31 pm
tonebeeze wrote:
Whitney Garner wrote:
anshumishra wrote: Thanks Whitney !
Although, I can't predict the right problem (that the intercept was to be -6 instead of 6 :) ), I definitely made a mistake when I didn't bother to check values of a and b. For sure, ab=k is a equation in 2-degree so, it is most likely to have a two values for a and b (which luckily converged to a single value here) . Conceptually, I was quite aware of it.

Thanks for pointing that out ! Hopefully, I don't miss it the next time.
and we definitely want to be careful that we don't mix up Y-intercepts with X-intercepts! It can be an easy mistake to make so we have to stay on our toes :)

Whit
I still don't fully understand problem. maybe it would help if someone would explain the problem to me without the abbreviations. I would really like to get this problem nailed down. thanks
tonebeeze,

Please follow either Whitney's solution or mine. I have already pointed in capital letter where I made the mistake. So, it would help you to go through the solution and fix the error(s) I made. Check the first post. This way hopefully you will get a better understanding.

Or, I am sure someone else will anyway solve it again.
Thanks
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by Whitney Garner » Mon Jan 03, 2011 9:30 pm
Hi tonebeeze!

So, first step is to correct the error in the question - it should read as follows:
  • 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)
We always want to start by rephrasing the question if possible. Points where a function intersect the x-axis are places where the y values are 0, so we can set y=0 to solve.

0 = (x+a)(x+b)

Which means that this graph will have x-intercepts at x+a=0 and x+b=0 ---> so when x= -a or -b. That means our rephrased question is "what are a and b?". One thing we should notice here is that we don't actually care what a or b are specifically because the letters are simply place holders. For example, the function y=(x+2)(x+3) would have the same intercepts as y=(x+3)(x+2) so it won't matter if a=6 and b=2, or b=6 and a=2 (so long as we are able to find a single pair of values).

Statement (1)
From this statement, there is no way to narrow down specific values for a and b nor to find a fixed pair. We can quickly test; if a=5, b=-6, but we could let a=0, and now b=-1. Because we have found 2 different pairs of values, this is INSUFFICIENT

Statement (2)
Knowing that the point (0,-6) is on the graph allows us to plug into the original function. We know that when x=0, y=-6 so we get:

-6 = (0+a)(0+b)
-6 = ab

But here again we cannot find a unique pair of numbers that multiply to -6. For example a=1, b=-6 or a=3,b=-2. INSUFFICIENT.

Statement (1+2 Together)
First we can rearrange the first expression to give a=-1-b. From here, we can substitute this into the equation from the second statement:

-6=(-1-b)*b
-6=-b - b^2
b^2 + b - 6 = 0
(b + 3)(b - 2) = 0
b=-3, 2

Here it would appear that we have 2 unique values for b, but remember, we only need a unique PAIR of values for a and b.

When b= -3, a= -1 - (-3) = -1+3 = 2, so the intercepts occur at x=-3, and x=2
When b= 2, a = -1 - (2) = -1 - 2 = -3, so the intercepts occur at x=2 and x=-3.

These are the SAME pairs...SUFFICIENT.

The answer is C.

Hope this clears it all up!
:)
Whit
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by goyalsau » Tue Jan 04, 2011 3:14 am
Whitney Garner wrote:Hi tonebeeze!

So, first step is to correct the error in the question - it should read as follows:
  • 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)
I am very puzzled by equation. As i know this is not a equation of circle not even straight line. Can you please explain what type of figure will be formed with this type of equation. Or we are not required to bother about the type of equation............
Saurabh Goyal
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-------------------------


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by anshumishra » Tue Jan 04, 2011 5:11 am
goyalsau wrote:
Whitney Garner wrote:Hi tonebeeze!

So, first step is to correct the error in the question - it should read as follows:
  • 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)
I am very puzzled by equation. As i know this is not a equation of circle not even straight line. Can you please explain what type of figure will be formed with this type of equation. Or we are not required to bother about the type of equation............
The equation is : y = x^2 + x(a+b) +ab

This is a quadratic equation which is also a parabola; this cuts the x-axis at (-a,0) and (-b,0)
Thanks
Anshu

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by goyalsau » Tue Jan 04, 2011 5:56 am
anshumishra wrote:
goyalsau wrote:
Whitney Garner wrote:Hi tonebeeze!

So, first step is to correct the error in the question - it should read as follows:
  • 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)
I am very puzzled by equation. As i know this is not a equation of circle not even straight line. Can you please explain what type of figure will be formed with this type of equation. Or we are not required to bother about the type of equation............
The equation is : y = x^2 + x(a+b) +ab

This is a quadratic equation which is also a parabola; this cuts the x-axis at (-a,0) and (-b,0)
How we know that it cuts parabola at -a,0 and -b,0.
Saurabh Goyal
[email protected]
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by anshumishra » Tue Jan 04, 2011 6:11 am
goyalsau wrote:
anshumishra wrote:
goyalsau wrote:
Whitney Garner wrote:Hi tonebeeze!

So, first step is to correct the error in the question - it should read as follows:
  • 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)
I am very puzzled by equation. As i know this is not a equation of circle not even straight line. Can you please explain what type of figure will be formed with this type of equation. Or we are not required to bother about the type of equation............
The equation is : y = x^2 + x(a+b) +ab

This is a quadratic equation which is also a parabola; this cuts the x-axis at (-a,0) and (-b,0)
How we know that it cuts parabola at -a,0 and -b,0.
y = (x+a)(x+b). This will cut x-axis when y=0 (y=0 is the equation for x-axis).
So, when x=-a, y =0 and when x=-b, y=0. {(-a,0) and (-b,0) is the point of intersection.}
Thanks
Anshu

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