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Source: — Data Sufficiency |
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by Tryingmybest » Sun Nov 23, 2008 10:44 am
y=(x+a)(x+b)
= x^2 -(a+b)x +ab --- 1

This intesects y axis at (o,-6)
0= 36 -(a+b)6 + ab

Now we should know a + b to solve the equation which is provided by A
0= 36 +6 + ab
=> ab=-42

So 1 becomes x^2 +x -42 = 0
(X-7) (X+6) =0
So intesection points will be ( 7,0) and (-6,0)

So C :)
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by sheena_gmat » Sun Nov 23, 2008 11:30 am
Tryingmybest wrote:y=(x+a)(x+b)
= x^2 -(a+b)x +ab --- 1

This intesects y axis at (o,-6)
0= 36 -(a+b)6 + ab

Now we should know a + b to solve the equation which is provided by A
0= 36 +6 + ab
=> ab=-42

So 1 becomes x^2 +x -42 = 0
(X-7) (X+6) =0
So intesection points will be ( 7,0) and (-6,0)

So C :)
isn't it sppose to be : x^2 + (a+b)* x + ab ?

Also x=0 and y=-6
You have taken the opposite
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by Tryingmybest » Sun Nov 23, 2008 12:19 pm
Yeah ....thanks for pointing out.
But still the solution is C.

y=(x+a)(x+b)
= x^2 +(a+b)x +ab --- 1

This intesects y axis at (o,-6)
=>-6=ab

Now we should know a + b to solve the equation which is provided by A


So 1 becomes x^2 -x -6= 0
(X-3) (X+2) =0
So intesection points will be ( -2,0) and (3,0)

So C
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by bhumika.k.shah » Wed Feb 03, 2010 7:41 am
But is it fine if we are getting two values for x ? that is -2 and 3 ??
Tryingmybest wrote:Yeah ....thanks for pointing out.
But still the solution is C.

y=(x+a)(x+b)
= x^2 +(a+b)x +ab --- 1

This intesects y axis at (o,-6)
=>-6=ab

Now we should know a + b to solve the equation which is provided by A


So 1 becomes x^2 -x -6= 0
(X-3) (X+2) =0
So intesection points will be ( -2,0) and (3,0)

So C
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by ajith » Wed Feb 03, 2010 8:04 am
bhumika.k.shah wrote:But is it fine if we are getting two values for x ? that is -2 and 3 ??
It is perfectly Fine, often some curves intersect the axes at 2 points [e.g. parabola]
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by bhumika.k.shah » Wed Feb 03, 2010 8:05 am
ajith wrote: [e.g. parabola]
Huh?????
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by ajith » Wed Feb 03, 2010 8:09 am
bhumika.k.shah wrote:
ajith wrote: [e.g. parabola]
Huh?????
Image

This is a parabola intersecting x axis at two points
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by ramsharma » Wed Feb 03, 2010 11:47 pm
bhumika.k.shah wrote:But is it fine if we are getting two values for x ? that is -2 and 3 ??

Question itself asks for two values of X "what two points does the graph"
RAM SHARMA
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by sanju09 » Thu Feb 04, 2010 12:17 am
gorsar 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)

OG answer is C.

Can anyone explaine why answer is C?
The two points where the graph of y = (x + a)(x + b) intersect the x-axis are the roots of the quadratic equation x^2 + (a + b) x + a b = 0. Hence, if (a + b) and a b are known, it's done.

(1) Insufficient.

(2) The graph intersects the y-axis when x = 0, put in quadratic and get a b = -6. Insufficient

Blended! Sufficient.

[spoiler]C[/spoiler]
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