Question

(please help!! fan and medal!! i don't know where to start!!)

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

\[
K(H) = -(H^2)+4 = -((x^2+1)^2)+4
\]Simplify.

Answer 2

I don't even know!

Answer 3

Then you can find roots.

Answer 4

\(\bf K(H) = -(H^2)+4 = -[(x^2+1)^2]+4\to
\begin{cases}
4-[(x^2+1)^2]
\\ \quad \\
\bf {\color{brown}{ a^2-b^2 = (a-b)(a+b)}}
\end{cases}\)

Answer 5

As people have pointed out K(H) means wherever you see an x in K, replace it with
x^2 +1
and then set the expression equal to zero.
you get
\[ -(x^2+1)^2 + 4 =0 \]
add -(x^2+1)^2 to both sides to get
\[ (x^2+1)^2 =4 \]
the next step is to take the square root of both sides.
what do you get ?

Answer 6

\[x ^{2}+1=2\]

Answer 7

almost, you get \( x^2 + 1= \pm 2\)
now you have to solve two equations
\( x^2 + 1=2 \\x^2 + 1=-2 \)

Answer 8

can you solve the first equation (you get two answers for x)

Answer 9

1 and -1?

Answer 10

yes. now do the second equation

Answer 11

you should get
\[ x^2 = -3 \\ x= \pm \sqrt{-3} \]
that is normally written using \( i = \sqrt{-1} \), as
\[ x= \pm i \sqrt{3} \]

Answer 12

i was about to ask that

Answer 13

we use the idea that
\[ \sqrt{a \cdot b} = \sqrt{a} \sqrt{b} \]
and the definition \( i = \sqrt{-1} \), to write
\[ \sqrt{-3} = \sqrt{-1 \cdot 3} = \sqrt{-1} \sqrt{3} \\ = i \sqrt{3} \]

Answer 14

so the answer would be a?

Answer 15

you have 4 solutions (did you forget about +1 and -1 ) ?

Answer 16

oh, right.

Answer 17

so the answer is choice e, right?

Answer 18

yes, choice E

Answer 19

thanks phi

Answer 20

you can check all 4 solutions. for example, when x=1 we first do
H(1)=(1^2)+1 = 1+1= 2

now do K(2) = -(2^2)+4 = -4 + 4 = 0 (it works)

here is x= -i sqr(3)
\[ H(-i \sqrt{3})=(-i \sqrt{3})^2+1 = -i \cdot -i \cdot 3+1= i^2\cdot 3 +1= -3 +1= -2\]
and
K(-2)= -(-2)^2 + 4 = - 4 + 4 = 0 (also works)

Answer 21

thanks