anonymous
  • anonymous
last time I'll ask, it's hard so DON'T click if you know you can't solve it! H(x)=(x^2)+1 and K(x)=-(x^2)+4. If K(H)=0, what are the roots/solutions? a. ±i√3 b. ±i c. ±2 d. ±1 e. ±1, ±i√3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
cmtboy2
  • cmtboy2
d. ±1 i think
anonymous
  • anonymous
please, please explain why..
anonymous
  • anonymous
i think it's e but i'm not so sure

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More answers

cmtboy2
  • cmtboy2
can it be to answer
anonymous
  • anonymous
@Hero
anonymous
  • anonymous
no, its only 1
cmtboy2
  • cmtboy2
@eskawaiidesu can it be two Answer or no
anonymous
  • anonymous
no, it can only be one
cmtboy2
  • cmtboy2
ok
cmtboy2
  • cmtboy2
um.... i wold go with C. But D. is also is the right answer to
anonymous
  • anonymous
okay, i'll see..
cmtboy2
  • cmtboy2
ok
cmtboy2
  • cmtboy2
go with c
cmtboy2
  • cmtboy2
what grad are you in
anonymous
  • anonymous
I'm in college :/
cmtboy2
  • cmtboy2
oh :)
cmtboy2
  • cmtboy2
i have to go bye
TrojanPoem
  • TrojanPoem
H(x)=(x^2)+1 and K(x)=-(x^2)+4 substitute H(x) instead of x to get K(H) K(H) = -(x^2+1)^2 + 4 K(H) = - (x^4 + 1 + 2x^2) + 4 K(H) = -x^4 - 1 - 2x^2 +4 = -x^4 -2x^2 + 3 Let K(H) = 0 X^4 + 2x^2 - 3 = 0 X^2 = 1 , x^2 = -3 Get the roots x = + or - 1 x = + or - sqrt(3)i I assume, you can use the quadratic formula -b +- sqrt(b^2 - 4ac)/2a Can't find the hard part you 're talking about..
anonymous
  • anonymous
thank you
TrojanPoem
  • TrojanPoem
Any time.

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