Gmat perp DS-2

Question

please explain this ..i dont have the OA.

jbs · Answer

When the given graph will intersect the x-axis, the value of y will be 0.

Therefore, we have 

(x+a)(x+b)=0

which means that either x=-a or x=-b will be the x-co-ordinate of the point of intersection.

We thus need to find the value of a & b. Statements 1 & 2 are two equations with 2 variables. We can thus get the values of a & b.
The two points of intersection will be (-a,0) & (-b,0).

 Important note : we get two pairs of values for a & b. However, the value of the x co-ordinate doesn't change in this case. The points are (3,0) and (-2,0).

sportyrizwan · Answer

Is the answer E

I get two values of a and two values of b

so total four points of intersection of the graph with X-axis

Damager · Answer

y=(x+a) (x+b)
y=(x^2 + ax + bx + ab)
y=(x^2 +x (a+b) + ab

from 1 (a+b)= -1 insuff
from 2 -6=0+0+ab insuff 

Togather

at point of intersection y=0
0=x^2 + (-1)x + (-6) = x= (-1 + sqrt (1+24))/2 or (-1 - sqrt (1+24))/2

pmenon · Answer

I end up with E....

at the end, all i know is the two values are some -3 and +2 .... but we dont know which one a is, and which one b is

jbs · Answer

I don't think anyone read the  Important Note  I wrote in my post.

I will explain again.

Now we know that for the points of intersection at the x-axis , y=0.
Therefore,from the given equation in the question,
we have (x+a)(x+b)=0.

You do get 2 values for 'a' & 'b'.  However , the co-ordinates of the points of intersection do not change. You get exactly two possible co-ordinates.

The two possible values of 'a' are -3 & 2.
The two possible values of 'b' are 2 & -3.

i.e either  a= -3 & b=2 
OR
  a=2 & b=-3. 

Consider  scenario 1 . 
if a=-3 & b=2, we have 
(x-3)(x+2)=0. (since y=0)
Therefore x=3 OR x=-2.
Therefore the points of intersection would be  (3,0) and (-2,0) 

Consider  scenario 2 .
 if a=2 & b=-3,we have
(x+2)(x-3)=0
Therefore x=-2 OR x=3
Therefore the points of intersection would again be  (3,0) and (-2,0) 

Please note that the actual values of 'a' & 'b' are irrelevant. The question asks for the 2 points of intersection and we can derive them from the given information.

GMATBLACKBELT · Answer

y=(x+a)(x+b)

We are looking for point (x,0)

1: a+b=-1. No help here.

2: Y int: (0,-6) No help here either b/c

-6=(x+a)(x+b) plug in for x ---> ab=-6. We cannot find out the x int. w/ this.

1&2: y=x^2+xa+xb+ab -- > x(x+a+b) -6 ---> x(x-1)-6 --> x^2-x-6

(x-3)(x+2) the line crosses the x axis at either point 3 or -2. Insuff.


E

pmenon · Answer

you know what, after reading it again this morning, i change my answer to C.

Jbs is right in his post above; using both statements we can come up with values for a and b, but really you are ending up with just two points, (-3,0) and (2,0)

LM · Answer

y = (x+a)(x+b) = x^2 + (a+b)x +ab This is the equation of right side up Parabola. ( Positive general quadratic equation ) y = ax^2 + bx + c ( general positive quadratic equation ) y = a(x – h)^2 + k, then the vertex is the point (h, k). This makes sense, if you think about it. The squared part is always positive (for a right-side-up parabola). y = ax^2 + bx + c, the vertex (h, k) is found by computing h = –b/2a, and then evaluating y at h to find k. k = (4ac – b^2) / 4a Just the background if someone wants to know fundamentals :wink:

Okkkkk so the equation we have is y = x^2 + (a+b)x +ab vertex (h,k) can be found h= -(a+b)/2 k=(4ab-(a+b)^2)/4 from (i) we get the vertex as ( -1/2, ab-1/4) As we do not know ab, we can't determine the point where it corsses the X axis. From (ii), we know the value of ab=-6 Combining (i) and (ii), we can definitely find the points where the parabola crosses the x-axis. ( I am not finding, but the points would be possibly ( not necessary ) coresponding points on the positive side and negative side of the X axis ) Therefore the answer is C ************************************************************* Second simple and layman's approach equation y=x^2+(a+b)x+ab from (i) y= x^2-x+ab For y=0, x= ab and also x = 1-ab as we don't know value of ab, we can't say what are the two values of x From(ii), we get ab=-6 From (i) and (ii), we can definitely find the values of x where the parabola instersects. x=-6 and x= 7 :wink:

Ferihere · Answer

Another vote for C, 
we know to points, either (-3;2) and (2;-3) one of them

 Ans: C

Gmat perp DS-2

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards