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Geometry Graph

Expert replies
by mpaudena » Tue Oct 20, 2009 5:54 pm
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).

Answer and explanation please. Thanks in advance.
Join the discussion
Source: — Data Sufficiency |

by sanjana » Wed Oct 21, 2009 1:17 am
IMO : C

What we need to know here is the point at which a line intersects the x axis the y coordinate of that point will be 0.

Hence we basically need to find a and b here coz when y=0
(x+a)(x+b)=0 so the 2 points will be (-a,0) and (-b,0)

Statement 1:
------------
a+b=-1

We can get more than 1 combination of a and b for which a+b=-1
Eg. -6,5 ; -3,2 etc
Insufficient.

Statement 2:
------------

The graph intersects y axis at (0,-6)
hence from the given equation
-6=(0+a)(0+b)
hence, ab=-6
again we can get more than one combination of a and b for which ab=-6

Combining 1 and 2
There is only 1 combination that satisfies

a+b=-1
and
ab=-6
and that is -3,2
Hence the 2 points are (-3,0) and (-2,0)

Hence C.
Join the discussion

by mpaudena » Wed Oct 21, 2009 6:30 am
sanjana wrote:IMO : C

What we need to know here is the point at which a line intersects the x axis the y coordinate of that point will be 0.

Hence we basically need to find a and b here coz when y=0
(x+a)(x+b)=0 so the 2 points will be (-a,0) and (-b,0)

Statement 1:
------------
a+b=-1

We can get more than 1 combination of a and b for which a+b=-1
Eg. -6,5 ; -3,2 etc
Insufficient.

Statement 2:
------------

The graph intersects y axis at (0,-6)
hence from the given equation
-6=(0+a)(0+b)
hence, ab=-6
again we can get more than one combination of a and b for which ab=-6

Combining 1 and 2
There is only 1 combination that satisfies

a+b=-1
and
ab=-6
and that is -3,2
Hence the 2 points are (-3,0) and (-2,0)

Hence C.
Perfection!
Join the discussion