Question

Please, i desperately need help. It's winter break, my teacher is not able to help because of vacation and no video or math tips that i know of is helping me right now. I really want to understand this!

Answer 1

The roots are -1+4i and -1 - 4i

so we know

x = -1+4i or x = -1 - 4i

-------------------------------------------------------

now add 1 to both sides for both equations

x+1 = -1+4i+1 or x+1 = -1 - 4i+1

x+1 = 4i or x+1 = -4i

-------------------------------------------------------

Then square both sides to get rid of the 'i' term

x+1 = 4i or x+1 = -4i

(x+1)^2 = (4i)^2 or (x+1)^2 = (-4i)^2

(x+1)^2 = 16i^2 or (x+1)^2 = 16i^2

(x+1)^2 = 16i^2

(x+1)^2 = 16(-1)

(x+1)^2 = -16

-------------------------------------------------------

making sense so far?

Answer 2

so @AlexandervonHumboldt2 I multiplied that and got 3x^2-15x. i must be doing something wrong :(

Answer 3

@jim_thompson5910 WOW yes it does!

Answer 4

I'm glad it does, let's keep going

Answer 5

Now expand out the left side


(x+1)^2 = -16

(x+1)(x+1) = -16

x(x+1)+1(x+1) = -16

x*x+x*1 + 1*x + 1*1 = -16

x^2 + x + x + 1 = -16

x^2 + 2x + 1 = -16

Then you add 16 to both sides

x^2 + 2x + 1 = -16

x^2 + 2x + 1+16 = -16+16

x^2 + 2x + 17 = 0

Answer 6

So, if you were to solve x^2 + 2x + 17 = 0 for x, you would get x = -1+4i or x = -1 - 4i

I recommend you use the quadratic formula to check that claim.

Answer 7

k

Answer 8

\[x=\frac{ -b \pm \sqrt (b ^(2)-4ac }{ 2a }\]
that is the quadratic equation, correct? @jim_thompson5910 and thank you!

Answer 9

if you meant to say

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

then yes, that is the quadratic formula

Answer 10

sorry, yes! that's what i was trying to say. many thanks!

Answer 11

no worries, you have to say b^{2} to mean \(\Large b^{2}\)

if you aren't sure how to type something up in the equation format, and you see someone else write it in the equation format, you can right-click on the equation they wrote and go to "show math as" and then go to "TeX commands"

Answer 12

Another way to cross check is the completing the square method: \[x^2 +2x +17=0\]\[(x^2+2x)+17=0\]\[(x+1)^2 +17-1=0\]\[(x+1)^2=-16\]\[x+1 = \pm \sqrt{-16}\]\[x=-1\pm \sqrt{-1}\sqrt{16}\]\[x=-1\pm 4i ~~\checkmark \]

Answer 13

i tried that tip, and it worked! i can't thank you enough :) @jim_thompson5910
@Jhannybean so that's how you check it, i see! thanks so much for the tips guys :) I can use these to find out how to work out my other problems

Answer 14

@jim_thompson5910 Let's try this by using Vieta's Formula! :)

Answer 15

So we can let \(\alpha = -1+4i\) and \(\beta = -1-4i\) ?

Answer 16

We are trying to get to: \(x^2 -(\alpha + \beta)x +\alpha\beta = 0\)

So the sum: \(\alpha + \beta = -1+4i -1-4i = -2 ....\) stuck :\

Answer 17

what is \(\Large \alpha*\beta\) equal to in this case?

Answer 18

\[\alpha\beta = (-1-4i)(-1+4i) = 1-4i+4i-16i^2 = 17\]

Answer 19

shortcut: \[\Large (a+bi)*(a-bi) = a^2 + b^2\]

Answer 20

is it really??? oh my gosh.

Answer 21

yes,

\[\Large (a+bi)(a-bi) = a(a-bi)+bi(a-bi)\]
\[\Large (a+bi)(a-bi) = a^2-abi + abi - b^2i^2\]
\[\Large (a+bi)(a-bi) = a^2 +0abi - b^2(-1)\]
\[\Large (a+bi)(a-bi) = a^2 + b^2\]

Answer 22

wait.. then \((a+b)(a-b) = a^2-b^2\) ?

Answer 23

the 'i's change it a bit though

Answer 24

oh, because \(i^2 = -1\)

Answer 25

yeah

Answer 26

nice! :D

Answer 27

So now I need to figure out the sum..\(\alpha +\beta\)

Answer 28

so,

\[\Large x^2 -(\alpha + \beta)x +\alpha\beta = 0\]

\[\Large x^2 -(-2)x +17 = 0\]

\[\Large x^2 + 2x +17 = 0\]

Answer 29

you already figured out alpha + beta

Answer 30

Ohh... \(+4i -4i = 0\) Haha, i see.

Answer 31

I thought for some reason we needed to substitute \(i=\sqrt{-1}\) in there.

Answer 32

Thanks jim!!!!!! If i could give you another medal for this I def would.

Answer 33

sure thing

Answer 34

\[(a+bi)(a-bi) = a^2 - (bi)^2 = \cdots \]

Answer 35

Oh that's another cool way to look at it!!

Answer 36

\[(\heartsuit + \clubsuit) (\heartsuit - \clubsuit) ~~ =~~\heartsuit^2 - \clubsuit^2 \]

Answer 37

\(\huge\color{green}{O.O~~~~
AWESOME :)}\)

Answer 38

\(\huge\color{green}{(O.O + *.*)(O.O - *.*)}\) \(\huge\color{red}{=}\) \(\huge\color{green}{O.O^2 - *.*^2}\)
shorry ::D :D