## anonymous one year ago 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.

1. 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?

2. anonymous

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

3. hartnn

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

4. hartnn

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

5. hartnn

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

6. hartnn

or (k+2)^2 > 4

7. anonymous

I will send a picture of my working out.

8. anonymous

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

9. anonymous

10. anonymous

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

11. hartnn

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

12. anonymous

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

13. anonymous

Oh I see!

14. 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

15. hartnn

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

16. hartnn

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

17. anonymous

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

18. hartnn

symbols switch when you multiply a negative on both sides

19. hartnn

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

20. anonymous

Ok, I think I understand now. Thank you very much! Your explanations really helped me!