Solve 5x2 = -30x - 65.

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Solve 5x2 = -30x - 65.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[5x^2=-30x-65 \\ 5x^2+30x+65=0\] 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
i tried that and i wound up with \[\frac{ 6\pm \sqrt{-16} }{ -2}\]
\[\text{ dividing both sides by 5} \\ x^2+6x+13=0\]

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\[x=\frac{-6 \pm \sqrt{6^2-4(1)(13)}}{2(1)} \\ =\frac{-6 \pm \sqrt{36-52}}{2}=\frac{-6 \pm \sqrt{-16}}{2} =\frac{-6 \pm \sqrt{-1(16)}}{2} \\ =\frac{-6 \pm \sqrt{-1} \sqrt{16}}{2}\] you are right so far
just need to find sqrt(16) and just recall how we rename sqrt(-1)
how did you get -1 in radical and 16 ?
so 4? sqrt -1?
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
wait would it be d?
\[=\frac{-6 \pm i (4)}{2} \\ =\frac{-6}{2} \pm i \frac{4}{2}\] 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.
oh so A i got it ! lol
can you help me with this one as well ? Solve -3x2 + 30x - 90 = 0.
try dividing both sides by -3 and use quadratic formula again
lol hold on a baby stole my pen and paper lol
oh no!
i got it back lol
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
ok
k im stuck on almost the last part im at \[\frac{ -10\pm \sqrt{220} }{ 2 }\]
let me check what you have so far one sec
\[-3x^2+30x-90=0 \\ \text{ divide both sides by -3 just \to make things easier \to play with } \\ x^2-10x+30=0 \\ x=\frac{-(-10) \pm \sqrt{(-10)^2-4(1)(30)}}{2(1)}\]
so -(-10)=10 and 100-120=-20
\[x=\frac{10 \pm \sqrt{-20}}{2}\]
now sqrt(-20)=sqrt(-1)*sqrt(20) but there is a perfect square factor in 20 sqrt(-20)=sqrt(-1)*sqrt(4)*sqrt(5)
\[x=\frac{10 \pm \sqrt{-1} \sqrt{4} \sqrt{5}}{2}\] see if you can simplify from there
is it this?x equals 10 plus or minus 2i square root of 5
all over 2 right ?
huh?
\[x=\frac{ 10 \pm i (2)\sqrt{5}}{2}\] you didn't say anything about the 2 on bottom
ohj yea
anyways you can separate the fraction and simplify a bit more
with the 10 and the 2?
(A+B)/C=A/C+B/C so from previous we have: \[x=\frac{10}{2} \pm i \frac{2}{2}\sqrt{5}\]
x equals 5 plus or minus i square root of 5?
yes since 10/2=5 and 2/2=1
can you help me with the rest xDDD lol
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
Solve 5x2 = -30x - 65.
that is the same as the first equation you put
but we can do it again if you didn't understand it
oh lol i put the wrong one cx
Solve x2 + 4x + 8 = 0.
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
you can skip to comparing your equation to ax^2+bx+c=0
which I think your problem isn't identifying a,b, and c
your problem is playing with the square root and separating the fraction
\[x^2+4x+8=0 \\ \text{ is the same as } 1x^2+4x+8=0 \\ \\ a=1,b=4,c=8 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(4) \pm \sqrt{4^2-4(1)(8)}}{2(1)} \\\] what do you get when you simplify this monster (and he isn't a bad looking monster; just a few teeth)
\[\frac{ -4\pm \sqrt{16-32} }{ 2 }\] so \[\frac{ -4\pm \sqrt{-16} }{ 2 }\] but im stuck now :/
beautiful so far
I actually simplify sqrt(-16) for you above
would i divide the -16 by 4?
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)=?
so would it be x equals negative 4 plus or minus 2 I
all over ?
2
oh wait and why 2?
sqrt(16)=4 not 2
cause it was all over two
minus what i said
\[x^2+4x+8=0 \\ \text{ is the same as } 1x^2+4x+8=0 \\ \\ a=1,b=4,c=8 \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(4) \pm \sqrt{4^2-4(1)(8)}}{2(1)} \\ \\ x=\frac{-4 \pm \sqrt{16-32}}{2} =\frac{-4 \pm \sqrt{-16}}{2} \\ x=\frac{-4 \pm 4i }{2} \\ \text{ now separate the fraction and simplify a bit more }\]
so \[-2\pm2i\]
yes
since -4/2=-2 and 4/2=2 so yep \[x=\frac{-4}{2} \pm \frac{4}{2}i =-2 \pm 2i\]
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
with peterpan remember those kids never get pizza
that is true but i would find a way to create it xD
thank you again!!
np :)

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