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by Brent@GMATPrepNow » Fri Oct 10, 2014 7:33 pm
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 (0,-6)
Target question: At which two points of the graph does y=(x+a)(x+b) intersect the x-axis?

IMPORTANT: Let's examine the point where a line (or curve) crosses the x-axis. At the point of intersection, the point is on the x-axis, which means that the y-coordinate of that point is 0. So, for example, to find where the line y=2x+3 crosses the x-axis, we let y=0 and solve for x. We get: 0 = 2x+3
When we solve this for x, we get x= -3/2.
So, the line y=2x+3 crosses the x-axis at (-3/2, 0)

Likewise, to determine the point where y = (x + a)(x + b) crosses the x axis, let y=0 and solve for x.
We get: 0 = (x + a)(x + b), which means x=-a or x=-b
This means that y = (x + a)(x + b) crosses the x axis at (-a, 0) and (-b, 0)
So, to solve this question, we need the values of a and b

Aside: y = (x + a)(x + b) is actually a parabola. This explains why it crosses the x axis at two points.

Now let's rephrase the target question...
REPHRASED target question: What are the values of a and b?

Statement 1: a + b = -1
There's no way we can use this to determine the values of a and b.
Since we can answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: The line intercepts the y axis at (0,-6)
This tells us that when x = 0, y = -6
When we plug x = 0 and y = -6 into the equation y = (x + a)(x + b), we get -6 = (0 + a)(0 + b), which tells us that ab=-6
In other words, statement 2 is a fancy way to tell us that ab = -6
Since there's no way we can use this information to determine the values of a and b, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
Statement 1 tells us that a+b = -1
Statement 2 tells us that ab = -6

Rewrite equation 1 as a = -1 - b
Then take equation 2 and replace a with (-1 - b) to get: (-1 - b)(b) = -6
Expand: -b - b^2 = -6
Set equal to zero: b^2 + b - 6 = 0
Factor: (b+3)(b-2) = 0
So, b= -3 or b= 2

When b = -3, a = 2 and when b = 2, a = -3
In both cases, the two points of intersection are (3, 0) and (-2, 0)
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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by [email protected] » Sat Oct 11, 2014 11:43 am
Hi kamalakarthi,

Like most questions on the GMAT, this question can be approached in a number of different ways. There's actually a great Algebra pattern/shortcut built into this question:

We given the equation Y = (X+A)(X+B) and we're asked at what 2 points the graph will intersect with the X-axis. This essentially comes down to the A and B. If we know their values, then we can answer the question. It's also worth noting that since we're multiplying, you can "flip-flop" the values of A and B and you'd have the same solution.

For example: (X+1)(X+2) is the same as (X+2)(X+1)......

Fact 1: A + B = -1

There's no way to determine the exact values for A and B with this information.

TESTing Values, we could have:
A = 0, B = -1
A = 100, B = -101
Etc.
Different numbers for A and B would lead to different solutions.
Fact 1 is INSUFFICIENT

Fact 2: The graph intersects the Y-axis at (0, -6).

Now we have one of the points on the graph. Plugging it into the original equation gives us....

-6 = (0+A)(0+B)
-6 = AB

We have the same situation as in Fact 1: more than 1 possible solution.
A = 1, B = -6
A = 2, B = -3
Etc.
Fact 2 is INSUFFICIENT

Combined, we have...
A + B = -1
AB = -6

This is a "system" of equations, so we CAN solve it, but we don't actually have to because the solutions would be flip-flops of one another.

If you did do the math, you'd have
A = -3, B = 2
OR
A = 2, B = -3

These provide the SAME solution.
Combined, SUFFICIENT

Final Answer: C

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by kamalakarthi » Sat Oct 11, 2014 12:34 pm
Thanks Rich. Just to get an idea on the difficulty, Can this question be considered at 650+ level.
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by Brent@GMATPrepNow » Sat Oct 11, 2014 1:38 pm
I'd say it's in the 650 - 750 range.

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by [email protected] » Sat Oct 11, 2014 6:03 pm
Hi kamalakarthi,

I agree with Brent's assessment about the general difficulty level of this question.

"Graphing" prompts on the GMAT are generally rare (1-2 is probably all that you'll see on Test Day) and generally far easier than this prompt (the given formula will lead you to graph a circle, which is really rare). This question is more of a conceptual question then a math or strategy question, although you could do math or use strategy (such as TESTing VALUES) to prove the correct answer. I wouldn't worry too much about this prompt, but some of the concepts behind it will certainly show up on other questions.

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by GMATGuruNY » Sat Oct 11, 2014 6:41 pm
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)
Alternate approach:

A graph intersects the x-axis when y=0.
Substituting y=0 into y = (x+a)(x+b), we get:
0 = (x+a)(x+b)
0 = x² + ax + bx + ab
0 = x² + (a+b)x + ab.

To determine for which values of x the right-hand side will be equal to 0, we need to know the values of the expressions in red (a+b and ab).

Statement 1: a+b = -1
Since the value of ab is unknown, INSUFFICIENT.

Statement 2: The graph intersects the y-axis at (0, -6)
Substituting (0, -6) into y = (x+a)(a+b), we get:
-6 = (0+a)(0+b)
-6 = ab.
Since the value of a+b is unknown, INSUFFICIENT.

Statements combined:
Since a+b = -1 and ab = -6, SUFFICIENT.

The correct answer is C.
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by KristenH88 » Sun Oct 12, 2014 5:06 pm
I don't get it What do you do after you've found a and b; that's where I stopped because I didn't see the correlation to the problem, and just chose E
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by Matt@VeritasPrep » Sun Oct 12, 2014 9:21 pm
KristenH88 wrote:I don't get it What do you do after you've found a and b; that's where I stopped because I didn't see the correlation to the problem, and just chose E
Kristen, we know that any point on the x-axis has a y-value of 0.

We know that y = (x + a)(x + b). Since we're touching the x-axis, y = 0. Since y and (x+a)(x+b) are the same thing, we know (x+a)(x+b) = 0.

Now we have our familiar quadratic equation. (x+a)(x+b) = 0 implies that x = -a or x = -b. So the question is really asking, "what are the values of -a and -b?" Once you have those, you're done!
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