Question

Find the range of values for x for which the equation x^2 - 2k(x+1) +k^2 = 6 has real roots. Find the roots in terms of k.

Answer 1

x^2 -2kx-2k+k^2-6=0
so sqrt(b^2-4ac) must be real
sqrt((2k)^2-4*(-2k+k^2)
so
((2k)^2-4*(-2k+k^2)>=0
4k^2+8k-4k-4k^2
4k>=0
so k>=0

Answer 2

thx!

Answer 3

wait a min

Answer 4

x^2 -2kx-2k+k^2-6=0
so sqrt(b^2-4ac) must be real
sqrt((2k)^2-4*(-2k+k^2) <-- where did the -6 go lol

Answer 5

@zzr0ck3r

Answer 6

good call, one sec

Answer 7

((2k)^2-4*(-2k+k^2-6)>=0
4k^2+8k-4k^2+24>=0
8k+24>=0
8k>=-24
k>=-3
OK im glad you saw that because my answer was not working when I try it.

Answer 8

and this one does when I check it:)

Answer 9

thanks!

Answer 10

find roots
x^2 -2kx-2k+k^2-6=0
(x-k)^2 = 2k-k^2+6+k^2
x = +_sqrt(2k-k^2+6+k^2)+k

Answer 11

and this is right

http://www.wolframalpha.com/input/?i=+%28sqrt%282k-k%5E2%2B6%2Bk%5E2%29%2Bk%29%5E2+-+2k%28%28sqrt%282k-k%5E2%2B6%2Bk%5E2%29%2Bk%29%2B1%29+%2Bk%5E2+%3D+

Answer 12

you see what I did?

Answer 13

I completed the square.

Answer 14

yup alright

Answer 15

should that say find the values of K for wich the equation has a solution?

Answer 16

which*

Answer 17

@BelleFlower ?

Answer 18

nope i copied the question correctly..

Answer 19

doesn't make sense.

Answer 20

that like
what values of x make x^2+2x+1=0 have real solutions
real solutions don't depend on the variable they depend on the constants. and K is the constand

Answer 21

constant*

Answer 22

I think it meant K, because what we did makes sense...

Answer 23

lol bye