Question

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.

Answer 1

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?

Answer 2

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

Answer 3

i though you said you used b^2 -4ac

what values of a,b,c did u get?

Answer 4

for 2 distinct points

b^2 -4ac must be > 0

Answer 5

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

Answer 6

or
(k+2)^2 > 4

Answer 7

I will send a picture of my working out.

Answer 8

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

Answer 9

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

Answer 10

oh yeah,
\((k+2)^2 -16 > 0 \\ (k+2)^2 >16\)

keep this as is and take square root on both sides...

Answer 11

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

Answer 12

Oh I see!

Answer 13

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

Answer 14

|k+2| > 4

k+2 > 4

or

k+2 < -4

Answer 15

it like if |a| >b

then a >b
or a < -b

Answer 16

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

Answer 17

symbols switch when you multiply a negative on both sides

Answer 18

in case of |a| > b

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