Data Sufficiency

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).

Deepthi Subbu 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).

when intercepting x-axis set y=0, x^2 +(a+b)*x + ab*x= 0 {foiling}

st(1) a+b=-1, x^2 +(-1)*x + ab*x= 0, can not solve for x, ab-? Not Sufficient;
st(2) y-intercept (0,-6) OR allowing distant line point as ab=-6 for the line equation in question OR x^2 +(a+b)*x + ab*x= 0 OR x^2 +(a+b)*x + (-6)*x= 0. can not solve for x, (a+b)-? Not Sufficient;
Combining st(1&2), x^2 +(-1)*x + (-6)*x= 0 OR x^2 - x + -6x = 0, x^2-7x=0 <=> x(x-7)=0 => x belongs {0;7} where x=0 for y-intercept and x=7 for x-intercept. Sufficient

Answer C

Deepthi Subbu 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).

The graph will intersect the x axis when y=0.
Since y = (x+a)(x+b), y=0 when either (x+a) = 0 or (x+b) = 0.
Thus, to determine which x values will make y=0, we need to know the values of a and b.

Statement 1:
2 variables, 1 linear equation, insufficient.

Statement 2:
Substituting x=0 and y=-6 into y = (x+a)(x+b), we get:
-6 = (0+a)(0+b)
-6 = ab.
2 variables, 1 equation, insufficient.

Statements 1 and 2 together:
2 variables, 2 different equations, sufficient.

The correct answer is C.

GMATGuruNY wrote:

Deepthi Subbu 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).

The graph will intersect the x axis when y=0.
Since y = (x+a)(x+b), y=0 when either (x+a) = 0 or (x+b) = 0.
Thus, to determine which x values will make y=0, we need to know the values of a and b.

Statement 1:
2 variables, 1 linear equation, insufficient.

Statement 2:
Substituting x=0 and y=-6 into y = (x+a)(x+b), we get:
-6 = (0+a)(0+b)
-6 = ab.
2 variables, 1 linear equation, insufficient.

Statements 1 and 2 together:
2 variables, 2 different linear equations, sufficient.

The correct answer is C.

Not necessarily (May be we are lucky here)

If,
a+b = k1 (say 5)
a*b = k2 (say 6) --- {Also, it is not linear }

We can get two pairs of values of a and b, in general [here, (2,3) and (3,2)].

anshumishra wrote:

GMATGuruNY wrote:

Deepthi Subbu 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).

The graph will intersect the x axis when y=0.
Since y = (x+a)(x+b), y=0 when either (x+a) = 0 or (x+b) = 0.
Thus, to determine which x values will make y=0, we need to know the values of a and b.

Statement 1:
2 variables, 1 linear equation, insufficient.

Statement 2:
Substituting x=0 and y=-6 into y = (x+a)(x+b), we get:
-6 = (0+a)(0+b)
-6 = ab.
2 variables, 1 equation, insufficient.

Statements 1 and 2 together:
2 variables, 2 different equations, sufficient.

The correct answer is C.

Not necessarily (May be we are lucky here)

If,
a+b = k1 (say 5)
a*b = k2 (say 6) --- {Also, it is not linear }

We can get two pairs of values of a and b, in general [here, (2,3) and (3,2)].

Yes, the values of a and b could be reversed. We would need to be concerned if:

-- there were a coefficient in front of either a or b in the linear equation
-- the problem asked for the value of a specifically or for the value of b specifically

For example, let's solve for the following 2 equations:

x+y = 7
xy = 12

Substituting y = 12/x into the first equation, we get:

x + 12/x = 7
x^2 + 12 = 7x
x^2 -7x + 12 = 0
(x-3)(x-4) = 0.
x = 3, 4.

Plugging these values into x+y=7, we get:
If x=3, y=4
If x=4, y=3.

Since there are no coefficients, only one combination of values will work: 3 and 4.

Now let's solve for a set of equations that includes a coefficient:

2x + y = 10.
xy = 12

Substituting y = 12/x into the first equation, we get:

2x + 12/x = 10
2x^2 + 12 = 10x
2x^2 - 10x + 12 = 0
(2x-4)(x-3) = 0.
x = 2, 3.

Plugging these values into 2x+y=10, we get:
If x=2, y=6
If x=3, y=4.

Because of the coefficient, more than one combination of values will work.

So the take-away is that given 2 equations, one of which gives the sum of 2 variables, the other of which gives the product, we'll have sufficient information if:

-- there are no coefficients
-- we need to know only what combination of values will satisfy the 2 equations

Since these 2 conditions are met in the DS question above, the 2 statements combined are sufficient, and I wouldn't waste time solving for a and b.

(Thanks for pointing out that ab = - 6 is not linear. Far from it, actually! I've amended my earlier post to ensure that no reader will be misinformed.)

GMATGuruNY wrote: So the take-away is that given 2 equations, one of which gives the sum of 2 variables, the other of which gives the product, we'll have sufficient information if:

-- there are no coefficients
-- we need to know only what combination of values will satisfy the 2 equations

I liked this part, which is right. For the current problem it works perfectly. Let me try to make it a bit clear by using an example to show when not to use it (or be careful):

What is point P(x,y) in a co-ordinate plane ?

(1) x+y = 7
(2) xy = 12

Even though:
-- there are no co-efficients
-- we need to know only what combination of values will satisfy the 2 equations

BUT, These two equations are satisfied by both (3,4) and (4,3), So the answer must be "E". I am posting it here, just so that one is careful about it. As long as you are are not trapped, PEACE !

ORIGINAL QUESTION - 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).

MY RESPONSE -
The first thing I wanted to do / did do was expand the equation to get
y = x^2 + bx + ax + ab
y = x^2 + x(a+b) + ab
0 = x^2 + x(a+b) + ab

Therefore in order to solve for the x intercept, we must make y = 0. We also need to know the values of both a and b.

Stmt 1: INSUFFICIENT
a + b = -1
Equation above now looks like this
0 = x^2 + x(-1) +ab
While this gives us (a+b) we still don't know ab

Stmt 2: INSUFFICIENT
(0,-6)
Plug these values into my above eqn i.e., x = 0, y = -6
-6 = 0^2 + 0(a+b) + ab
-6 = ab
So now our equation looks like this
0 = x^2 + x(a+b)-6

Now consider Stmt 1 + 2 together
This means we know both (a+b) and ab. Which would suggest C.
However, CAN SOMEONE PLEASE HELP ME THIS THIS - I know it's not necessary for the problem but when I actually try to solve for x I get stuck.....
0 = x^2 + x(-1) -6
0 = x^2 -x -6
0 = (x+2) (x-3)
Doesn't this mean that the x intercept could be either -2 or 3????? Or have I done something wrong. Does this mean really that the answer is E, as you we don't have a unique value for the x intercept?
Instead we have two possibilities
[-2,0] OR [3,0]

Can someone please explain where or what I've done wrong?

Thanks

Hi,reading closely the question, we see it says: In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis? ... the points are 2, actually (-2,0) and (3,0).

I hope it helps...

Bye

both together are sufficient but not alone.

(C) is answer
QED

ab=-6
a+b=-1

2 points can be found.

Answer C

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

we need both a+b and ab values

using both (1) and (2), it will be sufficient

Ans: C

C the answer.

a+b=-1
ab=-6

{a=2,b=-3}

{a=-3,b=2}

They are the same points which is (-3,2)

Statement 1: given a+b=-1
Statement 2: putting x=0 and y=6
we get ab=-6

We have two equations two variables
solving we get 2 solns for a and b
which are just the same and since the question doesnt specifically ask the value of a or b rather the two value
OA is C

C, a+b= -1 , ab=-6