graph intersects x-axis

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graph intersects x-axis

by shahab03 » Tue Feb 23, 2010 7:50 pm
Image


this question has been discussed quite a few times but I DON'T GET IT. Can someone please help this dumb person understand this:

I looked at following explanation but I dont understand why we need value for ab and why we are worried about (a+b). some background will be helpful..

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)
at x axis: y=0 ; x^2+(a+b)x+ab=0
1) a+b= -1 -insuffi
2) The graph intersects the y-axis at (0,-6); ==> ab=-6 -- insuff
both: suffi.
it is C...!!


From 1, a+b=-1.
From 2, x=0, so ab=6.
(x+a)*(x+b)=0
x^2+(a+b)x+ab=0
So, x=-3, x=2
The answer is C.

y = x^2 + (a+b)x + ab
A) a+b = -1..
But dont know ab
B) When x = 0, y = -6
Put in the 1st equation
We get, ab = -6
But dont know (a+b)
Combining C) y = x^2 -x -6
When it intersects x axis, y =0...
Thus, x^2-x-6 = 0.
The parabola cuts the x-axis at 3 and -2.
Hence, C

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by ldoolitt » Wed Feb 24, 2010 8:39 am
I'll give it a try...

You have an equation

y = (x+a) * (x+b)

What you want to know is the values of x that satisfy the quadratic equation

(k) 0 = (x+a) * (x+b)

Because, by definition, the x-intercept is the value of x in an equation when y is zero. This is what the question stem is asking for. Good so far?

Now, multiply (x) out and you get

(l) 0 = x^2 - x(a + b) + a*b

Now we can start looking at the data. From (1), we know that

a + b = -1

Note that in (l) we have an a + b! So plug in -1 for a + b in (l)

0 = x^2 - x*-1 + a*b
(m) 0 = x^2 +x + a*b

Now, can you plug that into the quadratic equation and get a number? No you can't because you don't know what a*b is. If you knew that a*b was, say, -20, you would have

0 = x^2 + x + -20
0 = (x + 5)(x - 4)

Which is an equation you can solve. But since you don't know what a*b is you don't know the solution. So (1) is not sufficient.

Lets look at (2). This gives you that there is a point on the line (0, -6). Lets plug that into the original equation

-6 = (0 + a)(0 + b)
(n) -6 = a*b

Unfortunately this information does not tell you anything about the value of x when y is zero, its giving you the value of y when x is zero. So we can see that (2) is not sufficient.

Looking at them both together, lets look at the equations labeled (m) and (n), which is what we are given from each of (1) and (2)

(m) 0 = x^2 +x + a*b
(n) -6 = a*b

But note that if you plug (n) into (m) you have

0 = x^2 + x - 6
0 = (x + 3)(x - 2)

So x can be either -3 or 2. Hence (1) and (2) together are sufficient and (c) is correct.

Helpful?