anonymous
  • anonymous
Use quadratic formula to solve for y 3x^2- 2(sqrt3)xy +y^2 +6(sqrt3)x -2y -40=0
Mathematics
schrodinger
  • schrodinger
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myininaya
  • myininaya
so we can put y in terms of x?
myininaya
  • myininaya
\[(1)y^2+(-2 \sqrt{3}x-2)y+(3x^2+6 \sqrt{3}x-40)=0\]
myininaya
  • myininaya
\[\text{ where }a=1, b=-2 \sqrt{3}-2 , c=3x^2+6 \sqrt{3} x-40\]

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myininaya
  • myininaya
\[y=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
well the option that is closest to what i get when i plug those in is y=1+sqrt3(x) +/- sqrt(-4sqrt3(x) =41)
myininaya
  • myininaya
\[y=\frac{-(-2 \sqrt{3}-2) \pm \sqrt{(-2 \sqrt{3}-2)^2-4(1)(3x^2+6 \sqrt{3}x-40)}}{2(1)}\]
myininaya
  • myininaya
this is crazy lookin' lol
myininaya
  • myininaya
\[y=\frac{2 \sqrt{3} +2 \pm \sqrt{4(3)-2(-2 \sqrt{3} (2))+2^2-12x^2-24 \sqrt{3} x+160}}{2}\]
anonymous
  • anonymous
haha yes i know! that is what i got when i plugged them in. However, i have multiple choice, and t hat is not an option. \[y= 1+ \sqrt{3}x \pm \sqrt{-4\sqrt{3}x +41}\]
anonymous
  • anonymous
this is the only similar option
myininaya
  • myininaya
\[y=\frac{2 \sqrt{3} +2 \pm \sqrt{12+8 \sqrt{3}+4-12x^2-24 \sqrt{3} x+160}}{2}\]
anonymous
  • anonymous
idk its not one of my options :(
myininaya
  • myininaya
\[y=\frac{2 \sqrt{3}+2 \pm \sqrt{-12x^2-24 \sqrt{3}x+176+8 \sqrt{3}}}{2}\] i'm still simplying i'm just showing my steps
myininaya
  • myininaya
simplifying*
anonymous
  • anonymous
oh okay! sorry!
myininaya
  • myininaya
\[y=\frac{2 \sqrt{3}+2 \pm \sqrt{4} \sqrt{-3x^2-6 \sqrt{x} +44+2 \sqrt{3}}}{2}\]
myininaya
  • myininaya
\[y=\frac{2 \sqrt{3} +2 \pm 2 \sqrt{-3x^2-6 \sqrt{x}+44+2 \sqrt{3}}}{2}\]
myininaya
  • myininaya
notice each term has the common factor 2
myininaya
  • myininaya
\[y=\frac{ \sqrt{3}+1 \pm 1 \sqrt{-3x^2-6 \sqrt{x}+44+2 \sqrt{3}}}{1}\] \[y= \sqrt{3}+1 \pm \sqrt{-3x^2-6 \sqrt{x}+44+2 \sqrt{3}}\]
myininaya
  • myininaya
oops i'm missing x in my b term
myininaya
  • myininaya
:(
myininaya
  • myininaya
i have to do this on paper lol
myininaya
  • myininaya
can you give me a sec?
anonymous
  • anonymous
its alright! of course i can! do you think when you are done you could help me with another similar problem?
myininaya
  • myininaya
ok i got it
myininaya
  • myininaya
i'm gonna scan it
anonymous
  • anonymous
ok:)
myininaya
  • myininaya
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myininaya
  • myininaya
which is what you thought gj
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anonymous
  • anonymous
yes! thank you so much
myininaya
  • myininaya
that is weird the file posted twice
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myininaya
  • myininaya
how many times do you see the file?
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myininaya
  • myininaya
it keeps posting :(
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anonymous
  • anonymous
like 5 at least haha
myininaya
  • myininaya
ok i have to do something post your new question on the left ok and if i get back i will look at it
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anonymous
  • anonymous
okay! i will.
myininaya
  • myininaya
:) fun problem by the way
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anonymous
  • anonymous
haha yes it was! thanks for your help!

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