Quadratics help

[spiderman]

Hey guys, pls cud any1 help me wit this question......

Find the set of values of k for which the line y=kx-4 intersects the curve y=x^2-2x at two distinct points???

[astarmathsandphysics]

When i get home.

[staceyboy3]

y=kx-4 and y=x^2-2x

intersects so kx-4=x^2-2x

simplify into the form: x^2 - (k+2)x + 4 = 0

when it comes to points of intersection, you look a the discriminant D=b^2-4ac, when ax^2+bx+c=0
intersects at 2 points so D > 0

(k+2)^2 - 4*1*4 > 0
k^2 + 4k -12 > 0

Change it to k^2 + 4k -12 = 0
and solve for k

k=2 or k=-6

Now you draw a graph (parabola shape of U on x-axis)
Looking back at k^2 + 4k -12 > 0, we want all the values that are above the x-axis (>0)

       |                  |
        \                /       +
          \            /
--------0------0---------->
        -6  \ _ _ / 2          -

That occurs when k<-6 or k>2

Hope the diagram isn't too confusing 

[vanibharutham]

I do a different method

if this helps you here is how i do it:

y=kx-4 and y=x^2-2x

intersects so kx-4=x^2-2x

simplify into the form: x^2 - (k+2)x + 4 = 0

when it comes to points of intersection, you look a the discriminant D=b^2-4ac, when ax^2+bx+c=0
intersects at 2 points so D > 0

(k+2)^2 - 4*1*4 > 0
k^2 + 4k -12 > 0

Change it to k^2 + 4k -12 = 0
and solve for k

k=2 or k=-6

The same upto there

now, i draw a number line

<-------[-6]-----------------------[2]--------------->

To find out your range you pick values from the number line

lets say 0, and we plug this into our equation...

k² + 4k -12 > 0
(0) + (0) - 12 = - 12 which is NOT greater than zero

hence, we know that k can not lie between -6 and 2

so the only possibility is that

k < -6 and k > 2

[spiderman]

Thanks a lot guys!!!!!

[staceyboy3]

No problem