Question

H(x) with two complex solutions (completing the squares) please help

Answer 1

Are we solving a function, or do you want us to make up a function?

Answer 2

Makes a function @amorfide

Answer 3

its as simple as

\[H(x) = x^2 + 1\]

you can chek it by using the discriminant

Answer 4

So the solution is (x+I)(x-I)? @ campbell_st

Answer 5

No. you are thinking about difference of two squares

if you were to have
H(x)= x+2x+2
complete the square would be

(x+1)^2 -1+2

Answer 6

nope the solution is

set H(x) = 0 and solve for x

so subtract 1 from both sides

\[x^2 = -1\]

then taking the square root of both sides

\[x = \pm \sqrt{-1}\]

which is a complex number and requires i^2 = -1 to simplify

Answer 7

so the next line becomes

\[x = \pm \sqrt{1 \times i^2}\]

after substituting... so now you can get your complex solution.

Answer 8

hope it makes sense

Answer 9

if you want a question that uses completing the square try

H(x) = x^2 +4x + 12

complete the square in x

H(x) = (x^2 + 4x + 4) + 8

then

H(x) = (x + 2)^2 + 8

set H(x) = 0

and

(x + 2)^2 + 8 = 0

subtract 8 from both sides

(x + 2)^2 = -8

take the square root of both sides

\[x + 2 = \pm \sqrt{-8}\]

subtract 2 from both sides

\[x = -2 \pm \sqrt{-8}\]

use i^2 = -1

substitute and simplify