Question

PLEASEEE HELP ME :( 1.)To open these doors, you must speak the function in standard form...
*One function, h(x), with two complex solutions.
....... i need an example of this im confused :(

Answer 1

hmm have you covered the quadratic formula?

Answer 2

... or do you know what a "complex number" is, and have you done quadratic equations yet?

Answer 3

yes i know that a complex number is a combination of a real and imaginary number, and yes.

Answer 4

ok... so you just need to make a quadratic equation with 2 complex solutions
so, the way to do it, will be to first, pick your 2 complex solutions, so which one do you think we can use?

Answer 5

got any 2 complex expressions we can use, pick any 2

Answer 6

sorry my computer froze

Answer 7

but is -6i and -12i???? work

Answer 8

sure

Answer 9

okay so what is the next step

Answer 10

ohh... one sec... yes so we know that the solutions are
x = -6i
x = -12i

so \(\bf x=-6i\implies x+6i=0\implies (x+6i)=0\\ \quad \\
x=-12i\implies x+12i=0\implies (x+12i)=0\\ \quad \\
\textit{so our polynomial will be } (x+6i)(x+12i)=0\)

Answer 11

i would not choose \(6i\) and \(12i\) because you will not get a polynomial with real coefficients
it is much easier than that

Answer 12

hmm

Answer 13

since \(x^2\geq 0\) for any \(x\) you can pick something like
\[x^2+4\] because \(x^2+4\geq 4\) and so has no zeros

Answer 14

ohhh... right...

Answer 15

so can i do x^2+4 and x^2+8

Answer 16

thank you all

Answer 17

yes you can

Answer 18

anyhow vickysecret , my bad, the pair usually will be a conjugate version of complex
so let's use say.... -6i and its conjugate +6i instead

\(\bf x=-6i\implies x+6i=0\implies (x+6i)=0\\ \quad \\
x=6i\implies x-6i=0\implies (x-6i)=0\\ \quad \\
\textit{so our polynomial will be } (x+6i)(x-6i)=0\\ \quad \\
\textit{recall that }(a-b)(a+b) = a^2-b^2\qquad thus\\ \quad \\
(x+6i)(x-6i)=0\implies x^2-(6i)^2=0\implies x^2-6^2(i^2)=0\\ \quad \\
x^2-36(-1)=0\implies x^2+36=0\)

Answer 19

okay that was alot, so is the official answer x^2+36=0

Answer 20

if we use -6i and +6i, yes
what you do is, from the solution, you get a polynomial, then multiply both polynomials to get the original function whose solutions are those

Answer 21

okay i understand now, thank you so much

Answer 22

yw