10. Find the set values of k for which the line y=kx-4 intersects the curve y=x^2-2x at two distinct points. Pleas help, I don't know what do do with this question. I tried solving it simultaneously and the using b^2-4ac but that didn't get me anywhere.

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10. Find the set values of k for which the line y=kx-4 intersects the curve y=x^2-2x at two distinct points. Pleas help, I don't know what do do with this question. I tried solving it simultaneously and the using b^2-4ac but that didn't get me anywhere.

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you tried? let me try the same thing, kx -4 = x^2 -2x x^2 - (k+2)x + 4 = 0 A = 1, B = -(k+2) , C = 4 you got same values?
Oh, no not at all. I didn't even end up getting any values.
i though you said you used b^2 -4ac what values of a,b,c did u get?

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for 2 distinct points b^2 -4ac must be > 0
[-(k+2)]^2 - 4 > 0
or (k+2)^2 > 4
I will send a picture of my working out.
sorry [{-(k+2)}^2-4*1*4]
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So after k^2>-4k-12 I didn't know what to do
oh yeah, \((k+2)^2 -16 > 0 \\ (k+2)^2 >16\) keep this as is and take square root on both sides...
\[\left( k+2 \right)^2>4^2,\left| k+2 \right|>4,k+2<-4,k<-6~or~k+2>4,k>2\]
Oh I see!
Ok I am getting the right numbers now (2 and -6) but the symbols are wrong. I got k>2 and k>-6 but it actually k>2 and k<-6
|k+2| > 4 k+2 > 4 or k+2 < -4
it like if |a| >b then a >b or a < -b
So the symbols switch when you take the square root of everything? Depending on if you are working with the positive or negative?
symbols switch when you multiply a negative on both sides
in case of |a| > b |dw:1442155794098:dw|
Ok, I think I understand now. Thank you very much! Your explanations really helped me!

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