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In the xy-plane, at what two points does the graph of

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In the xy-plane, at what two points does the graph of [#permalink] New post 07 Dec 2007, 06:43
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C
D
E

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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)


Can somebody show me how to solve this? thanks
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Re: PS: xy-plane [#permalink] New post 07 Dec 2007, 08:19
tarek99 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)


Can somebody show me how to solve this? thanks


The line intesects the x axis when y is zero. The problem is that you have to many variables.

Stmt 1 alone reduces the number of variables from 4 to 3 and is not suff.

Stmt 2 alone reduces the number of variales from 4 to 2 and is not suff.

Together you can use stmt 1 to write the equation in terms of y,x,b. You can then use stmt 2 to plug in y and x and get b. You then have the equation of the line and can find where it intersects the x axis.
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Re: PS: xy-plane [#permalink] New post 07 Dec 2007, 14:59
y = (x + a) (x + b)
from the question, we need intercept of x, so y = 0
(x + a) (x + b) = 0
x = -a or -b
1) a+b = -1
2) lines intersects y-axis at (0,6)
so -6 =(0+a) (0+b)
ab = -6

buy solving one and two you can have x = 3, -2.
So we have two points and so I think ans is E or C???.
Pls let us know the OA.
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Re: PS: xy-plane [#permalink] New post 08 Dec 2007, 02:03
tarek99 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)


Can somebody show me how to solve this? thanks


I think it's C.

from st.1: a+b=-1 we know nothing about a and b
from st.2: a*b=-6. Also, more than one option

combine:
1) a=-3, b=2, so the intersect points is (-3;0) and (2;0)
2) a=2, b=-3. we get the same intersect points (-3;0) and (2;0)
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Re: PS: xy-plane [#permalink] New post 08 Dec 2007, 02:25
tarek99 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)


Can somebody show me how to solve this? thanks


E.

y = x^2 + x(a+b) +ab
To find the point on x axis, put y=0
x^2 + x(a+b) +ab = 0

Stat 1: x^2 - x +ab = 0; insuff
Stat 2: x^2 + x(a+b) -6 = 0; insuff.

Together:
a = 3 & b = -2 or a = -3 & b = 2

If a = 3 & b = -2 then,
x^2 + x - 6 = 0
x = 2 or -3

If a = -3 & b = 2 then,
x^2 - x - 6 = 0
x = -2 or 3

So we have 2 pairs of points of intersection:
Either (2,0) & (-3,0) or (-2,0) & (3,0)
Insuff.
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Re: PS: xy-plane [#permalink] New post 08 Dec 2007, 02:31
GK_Gmat wrote:
tarek99 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)


Can somebody show me how to solve this? thanks


E.

y = x^2 + x(a+b) +ab
To find the point on x axis, put y=0
x^2 + x(a+b) +ab = 0

Stat 1: x^2 - x +ab = 0; insuff
Stat 2: x^2 + x(a+b) -6 = 0; insuff.

Together:
a = 3 & b = -2 or a = -3 & b = 2

If a = 3 & b = -2 then,
x^2 + x - 6 = 0
x = 2 or -3

If a = -3 & b = 2 then,
x^2 - x - 6 = 0
x = -2 or 3

So we have 2 pairs of points of intersection:
Either (2,0) & (-3,0) or (-2,0) & (3,0)
Insuff.


you cannot have a=3 and b=-2, it will not satisfy st.1: a+b=-1, 3-2=1!
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 [#permalink] New post 08 Dec 2007, 08:13
guys, the OA to this is C. So we need to figure out how exactly is it so. would anyone please explain?
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Re: PS: xy-plane [#permalink] New post 08 Dec 2007, 09:42
elgo wrote:
tarek99 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)


Can somebody show me how to solve this? thanks


I think it's C.

from st.1: a+b=-1 we know nothing about a and b
from st.2: a*b=-6. Also, more than one option

combine:
1) a=-3, b=2, so the intersect points is (-3;0) and (2;0)
2) a=2, b=-3. we get the same intersect points (-3;0) and (2;0)


Elgo, how did you narrowed it down to -3, 2, plugged in?
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 [#permalink] New post 08 Dec 2007, 11:01
tarek99 wrote:
guys, the OA to this is C. So we need to figure out how exactly is it so. would anyone please explain?


I could try :)

With y=(x+a)*(x+b), the intersection with X axis is obtained by using the equation of the x axis.

The X axis has y=0 for equation.

So, now, we have the following system:
o y=(x+a)*(x+b)
o y=0

That means:
(x+a)*(x+b) = 0

Before going further and keeping a concrete view of we talk about, I would like to bring your attention on y=(x+a)*(x+b). It's the equation of parabola such that y = u*x^2 + v*x + w where u = 1, making it in the shape of a valley in the XY plane.

Having said that, we have 2 cases to make (x+a)*(x+b) = 0:
> x = -a
or
> x = -b

So, we know that the 2 points are at (-a,0) and (-b,0).

All we have to do now is to look if we can have values for a and for b.

(C) it is :)... We have to solve a symetrical system, from which a and b can be interverted yet we can know the 2 points (a and b are not the final things we are asked to know) :
o a+b = -1
and
o a*b = -6
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 [#permalink] New post 08 Dec 2007, 13:42
guys, i understand why A and B aren't suff., however, i'm still confused with how option C is suff. can somebody show me the exact steps of the calculation in order to make option C suff?
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 [#permalink] New post 08 Dec 2007, 14:12
tarek99 wrote:
guys, i understand why A and B aren't suff., however, i'm still confused with how option C is suff. can somebody show me the exact steps of the calculation in order to make option C suff?


(x+a)*(x+b)
= x^2 + (a+b)*x + a*b

Stat 1.... Give us a+b = -1

Stat 2... If x = 0, then y = -6....

So, y = -6 = x^2 + (a+b)*x + a*b = 0^2 + (a+b)*0 + a*b
<=> a*b = -6

Stat 1 & 2
o a + b = -1
o a*b = -6

U can solve it by replacing a or b in a*b = -6... But in the G day, u better not do so :)

U have 2 different equations with 2 unknown variables... and a symetrical system here from which we could not define what is a or b in certitude. Again, that's not relevant to conclude :)

Let me develop it anyway :)
o b = -1 - a = -(a+1)
o a*b = a*(-1*(a+1)) = -6
<=> -a*(a+1) = -6
<=> a^2 + a = 6
<=> a^2 + a - 6 = 0

Delta = b^2 - 4*a*c = 1 - 4*1*(-6) = 25 = (5)^2

Thus:
o a(1) = (-b - sqrt(Delta)) / (2*a) = 1/2 * (-1 - 5) = -3
o a(2) = (-b + sqrt(Delta)) / (2*a) = 1/2 * (-1 + 5) = 2

Then,
o b(1) = -(1+a(1)) = -(1-3) = 2
o b(2) = -(1+a(2)) = -(1+2) = -3

Finally, we have:
o a(1) = -3 and b(1) = 2
or
o a(2) = 2 and b(2) = -3

:)
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Re: PS: xy-plane [#permalink] New post 10 Dec 2007, 03:46
elgo wrote:
GK_Gmat wrote:
tarek99 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)


Can somebody show me how to solve this? thanks


E.

y = x^2 + x(a+b) +ab
To find the point on x axis, put y=0
x^2 + x(a+b) +ab = 0

Stat 1: x^2 - x +ab = 0; insuff
Stat 2: x^2 + x(a+b) -6 = 0; insuff.

Together:
a = 3 & b = -2 or a = -3 & b = 2

If a = 3 & b = -2 then,
x^2 + x - 6 = 0
x = 2 or -3

If a = -3 & b = 2 then,
x^2 - x - 6 = 0
x = -2 or 3

So we have 2 pairs of points of intersection:
Either (2,0) & (-3,0) or (-2,0) & (3,0)
Insuff.


you cannot have a=3 and b=-2, it will not satisfy st.1: a+b=-1, 3-2=1!


Missed that! Thanks Elgo! Answer should be C then.
Re: PS: xy-plane   [#permalink] 10 Dec 2007, 03:46
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