Gprep DS

[Fiver]

Need a quick method to solve this. Request all to discuss their approach.

[abhasjha]

IMO - C

The graph is y = (x+a) (x+b)

To find out at what point this graph intersect x-axis approach as follows:-

When a graph intersect x-axis, the intersection point would be (x,0).
When a graph intersect y-axis, the intersection point would be (0,y).


Applying the same to equation , we have
=> (x+a) (x+b) = 0
=> x^2 + (a+b)x + ab = 0
The two points can be determined by solving this equation or by finding out the values of a and b.


(1) a+b = -1 , we still dont know the values of a and b. INSUFFICIENT

(2) The graph intersect the y-axis at (0,-6).
So the graph y = (x+a) (x+b) can be rewritten as
-6 = (0 +a) (0+b)
-6 = ab. we still dont know the values of a and b. INSUFFICIENT



Combing (1) and (2) , we have a + b = -1 and ab = -6 ,

The two points would be x^2 + (a+b)x + ab = 0
x^2 -x -6 =0
x^2 -3x +2x -6 =0
x(x-3) +2(x-3) =0
(x-3) (x+2) = 0


So two points where the graph intersect is (3,0) and (-2,0). Hence SUFFICIENT using (1) and (2) so C.

[StudyGMAT]

The answer is C

[clock60]

agree with C
y=(x+a)(x+b) if y=0, (intersecting the x-line)
(x+a)(x+b)=0
x=-a
x=-b
find the value of a,b??
(1) a*b=-6 insufficient plenty meanings are possible..
(2) x=0,y=-6
so insert the value of x,y is given equation and as a result
a*b=-6 insufficient
together
ab=-6
a+b=-1,a=-1-b
(-1-b)*b=-6
b^2+b-6=0
b1=-3, b2=2
from this part i start to think that the answer is E as we have pair for b so we will have pair for a but in some cases pairs can duplicate each other and this is the main trick in this problem
b1=-3
a1=-1+3=2
so (2,-3)
b2=2
a2=-1-2=-3
so again (2,-3)
and required answer 2.-3 so C
Fiver thank you for good question!!

[Fiver]

Thanks Clock60 & abhasjha. OA is C.

[finsisher]

two equations two unknown answer has to be C