OpenStudy (anonymous):

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.

3 years ago
hartnn (hartnn):

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?

3 years ago
OpenStudy (anonymous):

Oh, no not at all. I didn't even end up getting any values.

3 years ago
hartnn (hartnn):

i though you said you used b^2 -4ac what values of a,b,c did u get?

3 years ago
hartnn (hartnn):

for 2 distinct points b^2 -4ac must be > 0

3 years ago
hartnn (hartnn):

[-(k+2)]^2 - 4 > 0

3 years ago
hartnn (hartnn):

or (k+2)^2 > 4

3 years ago
OpenStudy (anonymous):

I will send a picture of my working out.

3 years ago
OpenStudy (anonymous):

sorry [{-(k+2)}^2-4*1*4]

3 years ago
OpenStudy (anonymous):

3 years ago
OpenStudy (anonymous):

So after k^2>-4k-12 I didn't know what to do

3 years ago
hartnn (hartnn):

oh yeah, $$(k+2)^2 -16 > 0 \\ (k+2)^2 >16$$ keep this as is and take square root on both sides...

3 years ago
OpenStudy (anonymous):

$\left( k+2 \right)^2>4^2,\left| k+2 \right|>4,k+2<-4,k<-6~or~k+2>4,k>2$

3 years ago
OpenStudy (anonymous):

Oh I see!

3 years ago
OpenStudy (anonymous):

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

3 years ago
hartnn (hartnn):

|k+2| > 4 k+2 > 4 or k+2 < -4

3 years ago
hartnn (hartnn):

it like if |a| >b then a >b or a < -b

3 years ago
OpenStudy (anonymous):

So the symbols switch when you take the square root of everything? Depending on if you are working with the positive or negative?

3 years ago
hartnn (hartnn):

symbols switch when you multiply a negative on both sides

3 years ago
hartnn (hartnn):

in case of |a| > b |dw:1442155794098:dw|

3 years ago