IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: spiderman on November 27, 2009, 02:09:18 pm
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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??????
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When i get home.
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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 :)
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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
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Thanks a lot guys!!!!! :)
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No problem ;)