Find a quadratic equation with roots -1+4i and -1-4i @TheSmartOne

Question

thesmartone · Answer

Well If the solution is 1 and 2
Then it is
(x-1)(x-2)=0

So for this one can you guess from here?

haleyelizabeth2017 · Answer

Isn't there only supposed to be like (x-_)(x-_) because if I solve it like that, then there is gonna be (x-_+_)(x-_-_).......so I am completely lost on this :(

thesmartone · Answer

Yes.

thesmartone · Answer

So
$$(x+(1-4i))(x+(1+4i))$$

haleyelizabeth2017 · Answer

oh....okay.....

haleyelizabeth2017 · Answer

so do I just do something like FOIL to find the equation?

thesmartone · Answer

Yes...

thesmartone · Answer

And remember$$i^2=\sqrt{-1}$$

campbell_st · Answer

you could use 

$$x = -1 \pm \sqrt{-16}$$

and work backwards

haleyelizabeth2017 · Answer

$i^2$ is equal to -1 not $$\sqrt{-1}$$ I thought......:(

thesmartone · Answer

oops yes.

campbell_st · Answer

then $$(x + 1) = \pm \sqrt{-16}$$

square both sides

haleyelizabeth2017 · Answer

Okay.....sorry @campbell_st! I think I'm gonna stick with TheSmartOne's method....thank you though

haleyelizabeth2017 · Answer

so anyhoo.....how do I do it when (1-4i) is in parenthesis? I am so sorry but I am not getting math very much tonight....:(

thesmartone · Answer

$$(x+1-4i)(x+1+4i)=0$$

campbell_st · Answer

ok... if you square both sides you get

$$(x + 1)^2 = -16$$

all you need to know is that 

$$\pm 4i = \pm \sqrt{-16}$$

good luck

haleyelizabeth2017 · Answer

so is it just like FOIL but with 3 different ones instead of two?

campbell_st · Answer

yes... and your diffuclty will be the expansion.

thesmartone · Answer

We would get 
$$x^2+x+4ix+x+1+4i-4ix-4i-4i^2=0$$

haleyelizabeth2017 · Answer

okay thats what I got on paper.....then simplify to $x^2+2x+1-36i$ because the +4ix and -4ix cancel out......

haleyelizabeth2017 · Answer

whoops! I got that wrong......

thesmartone · Answer

Oh oops.

haleyelizabeth2017 · Answer

so $x^2+2x+1-20i-16i^2$

thesmartone · Answer

$x^2+x+4ix+x+1+4i-4ix-4i-16i^2=x^2+2x+1+8i-16i^2$

haleyelizabeth2017 · Answer

so that would actually be $x^2+2x+17-20i$ because -16*-1=+16 and that can combine (like terms) with +1

haleyelizabeth2017 · Answer

I am unsure what -20i would turn into so we could simplify it even further....:(

thesmartone · Answer

How did you get -20i ??

haleyelizabeth2017 · Answer

never mind......my mistake ha ha yeah......:(

haleyelizabeth2017 · Answer

sorry

thesmartone · Answer

my mistake too -4i+4i means no i

thesmartone · Answer

Let me go back and show you.. I made a few mistakes lol

haleyelizabeth2017 · Answer

so it is just $x^2+2x+17$ :D

thesmartone · Answer

$$(x+1-4i)(x+1+4i)=0$$
$$x^2+1x+4ix+1x+1+4i-4ix-4i-16i^2=0$$
$$x^2+4ix-4ix+1x+1x+4i-4i-16i^2=0$$
$$x^2+2x-16i^2=0$$

thesmartone · Answer

$\ i^2=-1$
So
$\ x^2+2x-16i^2=x^2+2x-16(-1)$
$\ x^2+2x+16$

haleyelizabeth2017 · Answer

+16? what happened to the +1?

thesmartone · Answer

Oops all these errors..

thesmartone · Answer

Lets backtrack.

haleyelizabeth2017 · Answer

haha I got it though.....you just forgot to add the 1 so it should be $x^2+2x+17$ :D

thesmartone · Answer

$\ x^2+1x+4ix+1x+1+4i-4ix-4i-16i^2=0$

$\ x^2+4ix-4ix+1x+1x+4i-4i-16i^2+1=0$

$\ x^2+2x-16i^2+1=0$

$\ x^2+2x-16i^2+1=x^2+2x-16(-1)+1$

$\ x^2+2x+17$

thesmartone · Answer

Finally

haleyelizabeth2017 · Answer

Thank you sooooo sooooo much for all the help :D you have NO idea how much I appreciate your help :) thanks again

thesmartone · Answer

No problem... Sorry for all the mistakes lol

haleyelizabeth2017 · Answer

hey, it's okay! We got the answer in the end still! :D

thesmartone · Answer

:)