anonymous
  • anonymous
Solve 5x2 = -30x - 65.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • 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
  • anonymous
i tried that and i wound up with \[\frac{ 6\pm \sqrt{-16} }{ -2}\]
myininaya
  • myininaya
\[\text{ dividing both sides by 5} \\ x^2+6x+13=0\]

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

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