anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zzr0ck3r
  • zzr0ck3r
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
anonymous
  • anonymous
thx!
anonymous
  • anonymous
wait a min

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anonymous
  • anonymous
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
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
good call, one sec
zzr0ck3r
  • zzr0ck3r
((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.
zzr0ck3r
  • zzr0ck3r
and this one does when I check it:)
anonymous
  • anonymous
thanks!
zzr0ck3r
  • zzr0ck3r
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
zzr0ck3r
  • zzr0ck3r
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+
zzr0ck3r
  • zzr0ck3r
you see what I did?
zzr0ck3r
  • zzr0ck3r
I completed the square.
anonymous
  • anonymous
yup alright
zzr0ck3r
  • zzr0ck3r
should that say find the values of K for wich the equation has a solution?
zzr0ck3r
  • zzr0ck3r
which*
zzr0ck3r
  • zzr0ck3r
@BelleFlower ?
anonymous
  • anonymous
nope i copied the question correctly..
zzr0ck3r
  • zzr0ck3r
doesn't make sense.
zzr0ck3r
  • zzr0ck3r
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
zzr0ck3r
  • zzr0ck3r
constant*
zzr0ck3r
  • zzr0ck3r
I think it meant K, because what we did makes sense...
zzr0ck3r
  • zzr0ck3r
lol bye

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