Question

Solve the equation.
x^2+ 8x + 20=0

Answer 1

@satellite73

Answer 2

ok nvm what i just said
the zeros are complex
is that ok?

Answer 3

yes I did. Basically I did, but I couldn't find what the factorials would be..

Answer 4

east to complete the square, above answer is incorrect, since \((x+5)(x+3)=x^2+8x+15\)

Answer 5

I just guessed (x+2) (x+10)=0 ... but i'm sure that's wrong.

Answer 6

\[x^2+8x+20=0\]
\[x^2+8x=-20\]
\[(x+4)^2=-25+16=-4\]
\[x+4=\pm\sqrt{-4}=\pm2i\]\[x=-4\pm 2i\]

Answer 7

it does not have real zeros
if you do not allow complex numbers in the class, then there are no solutions

Answer 8

(x+4)2=−25+16=−4 how's you get this part?

Answer 9

*how'd

Answer 10

i completed the square
did you get to that yet, or no?

Answer 11

WE learned about it awhile ago, but I forget...

Answer 12

and i made a mistake, the \(-25\) was a typo, it should have been \(-20+16=-4\) but the rest is right

Answer 13

are you ok up until
\[x^2+8x=-20?\]

Answer 14

cause the rest is real real easy
i can walk you through it if you like

Answer 15

yes I understand the first part if you could walk me through the rest that would be great. :)

Answer 16

ok sorry i stepped away
ready?

Answer 17

yes Iam and it's fine

Answer 18

ok we start with
\[x^2+8x=-20\] what is half of \(8\)?

Answer 19

4?

Answer 20

yes

Answer 21

so we are going to replace the left hand side by
\[(x+4)^2=-20+\] and we need to know what goes after the \(+\) sign on the right
what is \(4^2\)

Answer 22

16?

Answer 23

yes, and that is what goes on the right
so you turn
\[x^2+8x=-20\] in to
\[(x+4)^2=-20+16=-4\]

Answer 24

ohh okay I understand now. Thankyou so much!!!

Answer 25

the reason this works (not that you have do write this out) is that
\[(x+4)^2=x^2+8x+16\] so addes \(16\) to the left and therefore you have to add \(16\) to the right to compensate

Answer 26

* added

Answer 27

yw

Answer 28

oh okaay