## DS: Coordinates

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### DS: Coordinates

by haidgmat » Sun Nov 21, 2010 11:25 am
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)

OA: C

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by rishab1988 » Sun Nov 21, 2010 2:03 pm
IMO C

Here is why:

Question is where y intersects x axis. When y intersects x axis t will be 0 ( because the coordinate of the point will be (0,x))

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

This is a quadratic equation. We need a+b and ab to determine the value of x.

1) a+b=-1

x^2 -x +ab =0

now and b could be anything for eg a =2 b=-1 gives a+b=-1 and ab=-2

Whereas a=-1/2 and b=-1/2 gives ab =1/4

Hence we can't solve the quadratic.

Eliminate A and D

2) when x =0 y=-6

substitute this in prompt to get : -6=ab

Then x^2+(a+b)x+ab=0 becomes x^2+(a+b)x-6=0

since a=-2 and b=3 gives ab=-6 and a+b=1 and a=-6 and b=1 gives a+b=-5

We can't solve the quadratic.Hence insufficient

Eliminate B

Combining 1 and 2

ab=-6 a+b=-1

The quadratic becomes x^2-x-6=0

We can solve this equation.Hence sufficient. The question is about two points (if it were for 1 point.again insufficient).

What is the OA?

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by [email protected] » Thu Oct 17, 2019 8:27 am
haidgmat 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-axis at (0, -6)

OA: C
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

Cheers,
Brent

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### Re: DS: Coordinates

by GMATinsight » Tue Jun 23, 2020 4:42 am
CONCEPT: Intersections of Quadratic graph on x-Axis = roots of equation
Roots of teh given equation are , x = -1 and x = -b
To answer the question we need th evalues of a and b

STatement 1: a + b= -1
a = 0 and b = -1 or a = -2 and b = 1 (Inconsistent solutions hence
NOT SUFFICIENT

STatement 2: The graph intersects the y-axis at (0, -6)
i.e y-intercept of teh graph = -6
but, y=(x+a)(x+b) = x^2 + (a+b)x + ab
i.e. ab = -6
NOT SUFFICIENT

COmbining the statements
a + b= -1 and ab = -6

We can deduce that a = -3 and b = 2 hence
SUFFICIENT

haidgmat wrote:
Sun Nov 21, 2010 11:25 am
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)

OA: C
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### Re: DS: Coordinates

by [email protected] » Mon Jun 20, 2022 3:26 am
haidgmat wrote:
Sun Nov 21, 2010 11:25 am
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)

OA: C
Solution:

The two points where the graph of y = (x + a)(x + b) intersects the x-axis are the two x-intercepts of the graph. To determine the x-intercepts, we solve for x when y = 0:

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

x = -a or x = -b

We see that if we can determine the values of a and b, then we can determine the two x-intercepts, namely, -a and -b.

Statement One Alone:

a + b = -1

Since there are many pairs of values that can add to -1, we can’t determine the exact values of a and b. Statement one alone is not sufficient.

Statement Two Alone:

The graph intersects the y-axis at (0, -6).

This means when x = 0, y = -6. That is:

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

-6 = ab

Like statement one, there are many pairs of values that can multiply to -6, so we can’t determine the specific values of a and b. Statement two alone is not sufficient.

Statements One and Two Together:

From the two statements, we see that a + b = -1 and ab = -6. If we let b = -6/a and substitute it into the first equation, we have:

a + (-6/a) = -1

a^2 - 6 = -a

a^2 + a - 6 = 0

(a + 3)(a - 2) = 0

a = -3 or a = 2

Although it seems there are two distinct values for a instead of one unique value, let’s look at the corresponding values of b also.

Case 1: If a = -3, b = -6/(-3) = 2.

Case 2: If a = 2, b = -6/2 = -3.

Recall that we are looking for the two x-intercepts, which have the values of -a and -b. In the first case, we have the two x-intercepts as 3 and -2 and in the second, we also have the same two x-intercepts, -2 and 3. Therefore, both statements together are sufficient to answer the question.