I can sort of make an educated guess conceptually, but I can't quite seem to work this one out on paper.
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 axes at (0,-6)
quadratic coordinate plane
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Target question: At which two points of the graph does y=(x+a)(x+b) intersect the x-axis?jjayhart wrote: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 axes at (0,-6)
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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Hi jjayhart,
While this DS question is certainly layered (and more complex than average), there is an algebra shortcut that you can use to save some time.
Brent's explanation for Fact1 and Fact2 correctly explain why each is INSUFFICIENT.
Combining facts, we have this:
Y = (X+A)(X+B)
A+B = -1
Point (0,-6) is on the graph.
Plugging in the point, we have this:
-6 = (A)(B)
When combined with the other equation, we have a 2V2E system:
-6 = (A)(B)
A+B = -1
You don't have to do any more math. With 2 variables and 2 unique equations, you CAN solve the system and answer the question.
Final Answer: C
This type of "system math" shortcut actually appears in a variety of circumstances, so be on the lookout for it (it can also appear as a "false system", so make sure that you've taken enough notes to be sure that it's an actual 2V2E).
GMAT assassins aren't born, they're made,
Rich
While this DS question is certainly layered (and more complex than average), there is an algebra shortcut that you can use to save some time.
Brent's explanation for Fact1 and Fact2 correctly explain why each is INSUFFICIENT.
Combining facts, we have this:
Y = (X+A)(X+B)
A+B = -1
Point (0,-6) is on the graph.
Plugging in the point, we have this:
-6 = (A)(B)
When combined with the other equation, we have a 2V2E system:
-6 = (A)(B)
A+B = -1
You don't have to do any more math. With 2 variables and 2 unique equations, you CAN solve the system and answer the question.
Final Answer: C
This type of "system math" shortcut actually appears in a variety of circumstances, so be on the lookout for it (it can also appear as a "false system", so make sure that you've taken enough notes to be sure that it's an actual 2V2E).
GMAT assassins aren't born, they're made,
Rich