Question

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

Answer 1

d. ±1 i think

Answer 2

please, please explain why..

Answer 3

i think it's e but i'm not so sure

Answer 4

can it be to answer

Answer 5

@Hero

Answer 6

no, its only 1

Answer 7

@eskawaiidesu can it be two Answer or no

Answer 8

no, it can only be one

Answer 9

ok

Answer 10

um.... i wold go with C. But
D. is also is the right answer to

Answer 11

okay, i'll see..

Answer 12

ok

Answer 13

go with c

Answer 14

what grad are you in

Answer 15

I'm in college :/

Answer 16

oh :)

Answer 17

i have to go bye

Answer 18

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..

Answer 19

thank you

Answer 20

Any time.