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anonymous
 4 years ago
Use quadratic formula to solve for y
3x^2 2(sqrt3)xy +y^2 +6(sqrt3)x 2y 40=0
anonymous
 4 years ago
Use quadratic formula to solve for y 3x^2 2(sqrt3)xy +y^2 +6(sqrt3)x 2y 40=0

This Question is Closed

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so we can put y in terms of x?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[(1)y^2+(2 \sqrt{3}x2)y+(3x^2+6 \sqrt{3}x40)=0\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\text{ where }a=1, b=2 \sqrt{3}2 , c=3x^2+6 \sqrt{3} x40\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{b \pm \sqrt{b^24ac}}{2a}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the option that is closest to what i get when i plug those in is y=1+sqrt3(x) +/ sqrt(4sqrt3(x) =41)

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{(2 \sqrt{3}2) \pm \sqrt{(2 \sqrt{3}2)^24(1)(3x^2+6 \sqrt{3}x40)}}{2(1)}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1this is crazy lookin' lol

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{2 \sqrt{3} +2 \pm \sqrt{4(3)2(2 \sqrt{3} (2))+2^212x^224 \sqrt{3} x+160}}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha 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
 4 years ago
Best ResponseYou've already chosen the best response.0this is the only similar option

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{2 \sqrt{3} +2 \pm \sqrt{12+8 \sqrt{3}+412x^224 \sqrt{3} x+160}}{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0idk its not one of my options :(

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{2 \sqrt{3}+2 \pm \sqrt{12x^224 \sqrt{3}x+176+8 \sqrt{3}}}{2}\] i'm still simplying i'm just showing my steps

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{2 \sqrt{3}+2 \pm \sqrt{4} \sqrt{3x^26 \sqrt{x} +44+2 \sqrt{3}}}{2}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{2 \sqrt{3} +2 \pm 2 \sqrt{3x^26 \sqrt{x}+44+2 \sqrt{3}}}{2}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1notice each term has the common factor 2

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=\frac{ \sqrt{3}+1 \pm 1 \sqrt{3x^26 \sqrt{x}+44+2 \sqrt{3}}}{1}\] \[y= \sqrt{3}+1 \pm \sqrt{3x^26 \sqrt{x}+44+2 \sqrt{3}}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1oops i'm missing x in my b term

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1i have to do this on paper lol

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1can you give me a sec?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its alright! of course i can! do you think when you are done you could help me with another similar problem?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1which is what you thought gj

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes! thank you so much

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1that is weird the file posted twice

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1how many times do you see the file?

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1ok i have to do something post your new question on the left ok and if i get back i will look at it

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1:) fun problem by the way

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha yes it was! thanks for your help!
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