Question

Solve 5x2 = -30x - 65.

Answer 1

\[5x^2=-30x-65 \\ 5x^2+30x+65=0\]
you could divide both sides by 5
and see if you end up with something factorable on the the left hand side
or you can just use quadratic formula

Answer 2

i tried that and i wound up with \[\frac{ 6\pm \sqrt{-16} }{ -2}\]

Answer 3

\[\text{ dividing both sides by 5} \\ x^2+6x+13=0\]

Answer 4

\[x=\frac{-6 \pm \sqrt{6^2-4(1)(13)}}{2(1)} \\ =\frac{-6 \pm \sqrt{36-52}}{2}=\frac{-6 \pm \sqrt{-16}}{2} =\frac{-6 \pm \sqrt{-1(16)}}{2} \\ =\frac{-6 \pm \sqrt{-1} \sqrt{16}}{2}\]
you are right so far

Answer 5

just need to find sqrt(16) and just recall how we rename sqrt(-1)

Answer 6

how did you get -1 in radical and 16 ?

Answer 7

so 4? sqrt -1?

Answer 8

a. x = -3 ± 2i
b.x = -3 ± 4i
c.x = -6 ± 2i
d.x = -6 ± 4i

These are the answers that I got and I'm so confused

Answer 9

wait would it be d?

Answer 10

\[=\frac{-6 \pm i (4)}{2} \\ =\frac{-6}{2} \pm i \frac{4}{2}\]
can you simplify?

Are you asking where I got sqrt(-16)
didn't you also get that for that one part? That is what your expression said.

Answer 11

oh so A i got it ! lol

Answer 12

can you help me with this one as well ?
Solve -3x2 + 30x - 90 = 0.

Answer 13

try dividing both sides by -3
and use quadratic formula again

Answer 14

lol hold on a baby stole my pen and paper lol

Answer 15

oh no!

Answer 16

i got it back lol

Answer 17

see how far you can get with the problem and if you can't get all the way through it post what you can get through

Answer 18

ok

Answer 19

k im stuck on almost the last part im at \[\frac{ -10\pm \sqrt{220} }{ 2 }\]

Answer 20

let me check what you have so far
one sec

Answer 21

\[-3x^2+30x-90=0 \\ \text{ divide both sides by -3 just \to make things easier \to play with } \\ x^2-10x+30=0 \\ x=\frac{-(-10) \pm \sqrt{(-10)^2-4(1)(30)}}{2(1)}\]

Answer 22

so -(-10)=10
and 100-120=-20

Answer 23

\[x=\frac{10 \pm \sqrt{-20}}{2}\]

Answer 24

now sqrt(-20)=sqrt(-1)*sqrt(20)
but there is a perfect square factor in 20
sqrt(-20)=sqrt(-1)*sqrt(4)*sqrt(5)

Answer 25

\[x=\frac{10 \pm \sqrt{-1} \sqrt{4} \sqrt{5}}{2}\]
see if you can simplify from there

Answer 26

is it this?x equals 10 plus or minus 2i square root of 5

Answer 27

all over 2 right ?

Answer 28

huh?

Answer 29

\[x=\frac{ 10 \pm i (2)\sqrt{5}}{2}\]
you didn't say anything about the 2 on bottom

Answer 30

ohj yea

Answer 31

anyways you can separate the fraction and simplify a bit more

Answer 32

with the 10 and the 2?

Answer 33

(A+B)/C=A/C+B/C

so from previous we have:
\[x=\frac{10}{2} \pm i \frac{2}{2}\sqrt{5}\]

Answer 34

x equals 5 plus or minus i square root of 5?

Answer 35

yes since 10/2=5 and 2/2=1

Answer 36

can you help me with the rest xDDD lol

Answer 37

i think i can help with one more
but I want to see you work it all the way to the end
and I will check it and we can make changes if needed

Answer 38

Solve 5x2 = -30x - 65.

Answer 39

that is the same as the first equation you put

Answer 40

but we can do it again if you didn't understand it

Answer 41

oh lol i put the wrong one cx

Answer 42

Solve x2 + 4x + 8 = 0.

Answer 43

since we already did this one
I will be a bit more detailed with the explanation:
\[5x^2=-30x-65 \\ \text{ I first added } 30x \text{ and } 65 \text{ on both sides } \\ 5x^2+30x+65=0 \\ \text{ I then notice all terms are divisible by 5} \\ \text{ So I divided both sides by 5} \\ x^2+6x+13=0 \\ \text{ comparing this to } ax^2+bx+c=0 \\ \text{ I identified } a=1, b=6,c=13 \\ \text{ so using the quadratic formula we have } \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(6)\pm \sqrt{(6)^2-4(1)(13)}}{2(1)} \\ x=\frac{-6 \pm \sqrt{36-52}}{2} \\ x=\frac{-6 \pm \sqrt{-16}}{2} \\ \text{ now } \sqrt{-16}=\sqrt{-1(16)}=\sqrt{-1} \sqrt{16}=i (4)=4i \\ \text{ so we have } \\ x=\frac{-6 \pm 4i}{2} \\ \text{ now seperating the fraction } \\ x=\frac{-6}{2}\pm\frac{4i}{2} \\ x=\frac{-6}{2}\pm\frac{4}{2}i \\ x=-3 \pm 2i\]

now I see you have another equation there
see if you can follow what I did here to solve your equation

Answer 44

you can skip to comparing your equation to ax^2+bx+c=0

Answer 45

which I think your problem isn't identifying a,b, and c

Answer 46

your problem is playing with the square root and separating the fraction

Answer 47

\[x^2+4x+8=0 \\ \text{ is the same as } 1x^2+4x+8=0 \\ \\ a=1,b=4,c=8 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(4) \pm \sqrt{4^2-4(1)(8)}}{2(1)} \\\]
what do you get when you simplify this monster (and he isn't a bad looking monster; just a few teeth)

Answer 48

\[\frac{ -4\pm \sqrt{16-32} }{ 2 }\]
so \[\frac{ -4\pm \sqrt{-16} }{ 2 }\]
but im stuck now :/

Answer 49

beautiful so far

Answer 50

I actually simplify sqrt(-16) for you above

Answer 51

would i divide the -16 by 4?

Answer 52

sqrt(16)=4
sqrt(-1)=i

you have sqrt(-16) though
sqrt(-16)
remember -16 is same as -1*16
so sqrt(-16)=sqrt(-1*16)
=sqrt(-1)*sqrt(16)
you should be able to identify sqrt(-1) as i
and you should be able to simplify sqrt(16)
since you know 4^2=16 (Where 4>0) then sqrt(16)=?

Answer 53

so would it be x equals negative 4 plus or minus 2 I

Answer 54

all over ?

Answer 55

2

Answer 56

oh wait and why 2?

Answer 57

sqrt(16)=4 not 2

Answer 58

cause it was all over two

Answer 59

minus what i said

Answer 60

\[x^2+4x+8=0 \\ \text{ is the same as } 1x^2+4x+8=0 \\ \\ a=1,b=4,c=8 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(4) \pm \sqrt{4^2-4(1)(8)}}{2(1)} \\ \\ x=\frac{-4 \pm \sqrt{16-32}}{2} =\frac{-4 \pm \sqrt{-16}}{2} \\ x=\frac{-4 \pm 4i }{2} \\ \text{ now separate the fraction and simplify a bit more }\]

Answer 61

so \[-2\pm2i\]

Answer 62

yes

Answer 63

since -4/2=-2
and 4/2=2
so yep
\[x=\frac{-4}{2} \pm \frac{4}{2}i =-2 \pm 2i\]

Answer 64

i think i got the rest <3 thank you . now i gotta fill out some more job apps :/ ugh growing up sucks wish there was such thing as never land

Answer 65

with peterpan
remember those kids never get pizza

Answer 66

that is true but i would find a way to create it xD

Answer 67

thank you again!!

Answer 68

np :)