Solve 5x2 = -30x - 65.

- anonymous

Solve 5x2 = -30x - 65.

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- myininaya

\[5x^2=-30x-65 \\ 5x^2+30x+65=0\]
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

- anonymous

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

- myininaya

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

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## More answers

- myininaya

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

- myininaya

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

- anonymous

how did you get -1 in radical and 16 ?

- anonymous

so 4? sqrt -1?

- anonymous

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

- anonymous

wait would it be d?

- myininaya

\[=\frac{-6 \pm i (4)}{2} \\ =\frac{-6}{2} \pm i \frac{4}{2}\]
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.

- anonymous

oh so A i got it ! lol

- anonymous

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

- myininaya

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

- anonymous

lol hold on a baby stole my pen and paper lol

- myininaya

oh no!

- anonymous

i got it back lol

- myininaya

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

- anonymous

ok

- anonymous

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

- myininaya

let me check what you have so far
one sec

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

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

- anonymous

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

- myininaya

all over 2 right ?

- anonymous

huh?

- myininaya

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

- anonymous

ohj yea

- myininaya

anyways you can separate the fraction and simplify a bit more

- anonymous

with the 10 and the 2?

- myininaya

(A+B)/C=A/C+B/C
so from previous we have:
\[x=\frac{10}{2} \pm i \frac{2}{2}\sqrt{5}\]

- anonymous

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

- myininaya

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

- anonymous

can you help me with the rest xDDD lol

- myininaya

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

- anonymous

Solve 5x2 = -30x - 65.

- myininaya

that is the same as the first equation you put

- myininaya

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

- anonymous

oh lol i put the wrong one cx

- anonymous

Solve x2 + 4x + 8 = 0.

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

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

- myininaya

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

- anonymous

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

- myininaya

beautiful so far

- myininaya

I actually simplify sqrt(-16) for you above

- anonymous

would i divide the -16 by 4?

- myininaya

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)=?

- anonymous

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

- myininaya

all over ?

- anonymous

2

- myininaya

oh wait and why 2?

- myininaya

sqrt(16)=4 not 2

- anonymous

cause it was all over two

- anonymous

minus what i said

- myininaya

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

- anonymous

so \[-2\pm2i\]

- myininaya

yes

- myininaya

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

- anonymous

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

- myininaya

with peterpan
remember those kids never get pizza

- anonymous

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

- anonymous

thank you again!!

- myininaya

np :)

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