In the xy-plane, at what two points does the graph y = (x + a)(x + b)

Question

In the xy-plane, at what two points does the graph  y = (x + a)(x + b)  intersect the x-axis?

(1)  a + b = -6

(2) The graph contains the point  (0, -7) .

Mona2019 · Answer

It intersects the X axis when y =0
So basically the question is asking the solution of the equation (X+a)(X+b)=0

Statement one is insufficient as a and b can take many different combinations like -2&-4, -10&4 ..

Statement 2 is insufficient because
Substituting X with 0 and y with -7
The equation now is ab =-7 
A,B can now be 7&-1 or -7&1 or -3.5*2 ..etc

Now combining 1&2 we will get unique combination of a and b.

So C IMO is the answer

Posted from my mobile device

lipikatiwari · Answer

IMO C

The graph will intersect x axis at points where y=0. Putting y=0 in the given equation of graph, we get:
(x+a)(x+b)=0
i.e. x= -a,-b
Hence, the graph will intersect x-axis at points (-a,0) and (-b,0)

1) a+b=-6
This doesn't give us any unique value of a and b. There are mainy possibilities: (-6,0),(-5,-1),(1,-7) etc which will satisfy this condition.
So, statement 1 is not sufficient

2) This graph contains point (0,-7). Putting this in the equation:
-7=(0+a)(0+b)
ab=-7
There can be many combinations that satisfy this condition: (1,-7),(-1,7)(0.5,-14) etc
Statement 2 is not sufficient

Now using both 1 and 2:
In 1, we know a+b=-6
So, b = -(6+a)
Now using this value of b in Statement 2's condition (where ab=-7)
-(6+a)a=-7
6a+a^2=7
a^2+6a-7=0
Solving this quadratic equation, we get:
a=1,-7
Putting values of 'a' in either equation:
If a=1 => b=-7 
AND if a=-7 => b=1

So for a=1 and b=-7, equation of the graph becomes y=(x+1)(x-7)
For a=-7 and b=1, equation of graph becomes y=(x-7)(x+1)

Both the equations are same, y=(x+1)(x-7)
Putting y=0 for x intercepts,
0=(x+1)(x-7)
Hence, the two points that will intersect the graph are (-1,0) and (7,0)

Since we used both statements, answer is C

Posted from my mobile device

Bunuel · Answer

Official Solution:

A graph intersects the x-axis when its y-coordinate is 0. Thus, the question is asking us to find the two values of  x  that make the equation  (x + a)(x + b) = 0 . In order for the expression on the left to equal 0, either  (x + a)  or  (x + b)  must equal 0. If  (x + a) = 0 , then  x = -a , and if  (x + b) = 0 , then  x = -b . Therefore, the two points we are looking for are  (-a, 0)  and  (-b, 0) . We can find the solutions if we can determine the values of  a  and  b .
 
Statement 1 tells us that  a + b = -6 . This is a linear equation with two variables, so we cannot solve for either one. For example, both  a = -3 ,  b = -3  and  a = -6 ,  b = 0  would work. Statement 1 is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice must be B, C, or E.
 
Statement 2 tells us that the graph contains the point  (0, -7) . We plug this point into the equation from the prompt:  -7 = (0 + a)(0 + b) = ab . Again, we have a single equation with two variables, so we cannot solve for either one. Statement 2 is NOT sufficient to answer the question. Eliminate answer choice B. The correct answer choice must be C or E.

Evaluating the statements together, we obtain two equations with two unknowns:  a + b = -6  and  ab = -7 . Since these are independent equations, we have enough information to answer the question.

Answer  choice C  is correct.

Although we do NOT need to find the actual points of intersection, let us do so for the purposes of illustration. There are two approaches that we can take to find  a  and  b .

First approach: Since the GMAT likes to use simple, round numbers, we look for round numbers that would fit these equations. The two integers factors of 7 are 1 and 7, and  1 + (-7) = -6 ; thus,  a  and  b  must be 1 and -7. We do not need to determine which value corresponds to which variable, because the question asks for the TWO points at which the graph intersects the x-axis.

Second method: We have two equations with two variables, so we can combine the equations. Solving for  b  in  a + b = -6  gives:  b = -a - 6 = -(a + 6) . Plugging this value into  ab = -7 , we get:  a(-(a + 6)) = -7 , so  -a^2 - 6a + 7 = 0 , or  a^2 + 6a - 7 = 0 . We can factor this equation as  (a -1)(a + 7) = 0 , which means that  a = 1  or  a = -7 . Then, plugging these values into  ab = -7  to solve for  b  gives  b = -7  or  b = 1 . Either  a  is 1 and  b  is -7 or vice versa, so the points  (-a, 0)  and  (-b, 0)  are  (-1, 0)  and  (7, 0) . Again, we do not need to determine which is which.

Once we find  a  and  b , we know the two points at which the graph of  y = (x + a)(x + b)  intersects the x-axis. The statements together are sufficient to answer the question.

Answer: C

KalyanChamarthi · Answer

1. What do we need to find out?
two points at which the graph interests the x-axis, 
which means we need to fine the values of x co-ordinates of those two points.

2. What do we know from the question stem 
we know : y = (x+a)(x+b) 
 graph intersects the x-axis ==> y = 0
therefore, (x+a)(x+b) = 0 ==> x^2 + x(a+b)+ ab

3. What else do we need ? 
(a+b) = ? 
ab = ?

FS 1: a + b = -6, but we still don't know ab= ? hence not sufficient, so you can eliminate answer choice AD

FS 2 : the graph touches y axis @ (0,-7) ==> x = 0 and y = -7 
therefore, 
(0+a)(0+b) = -7 ==> ab = -7, but we still do have a+b = ? from FS2 Alone, hence not sufficient, so you can eliminate answer choice B

Now let us combine : FS1 and FS2, we have a+b = -6 and ab = - 7 
so when you put these values in the original equations, you can find the value of x by solving the quadratic equation
x^2-6x-7 = 0, as mentioned you can solve for x.

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In the xy-plane, at what two points does the graph y = (x + a)(x + b)

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