Practice Exam 1 - GMAT Prep -

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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Thu May 26, 2016 2:17 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

Cheers,
Brent
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by Matt@VeritasPrep » Thu May 26, 2016 2:18 pm
When you intersect the x-axis, y = 0, so we want the two solutions to

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

or the values of both a and b (though we don't care which is which!)

S1:

a + b = -1, two variables, one equation, not sufficient

S2:

This says x = 0, y = -6 is one point on y = (x + a)*(x + b), so we have -6 = (0 + a)*(0 + b), or -6 = a*b. This is two variables, one equation again, so not sufficient

S1 + S2:

Two variables, two equations, so we're set. (a = 2, b = -3, or vice versa.)