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In the xy-plane, at what two points does the graph of [#permalink]
 07 Dec 2007, 07:43
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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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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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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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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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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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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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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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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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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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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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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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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.
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