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Mathematics 13 Online
OpenStudy (anonymous):

WIll award medal Consider the equation below. 9x2 + y2 + 9z2 – 6y – 90z + 225 = 0 reduce the equation to one of the standard forms

OpenStudy (anonymous):

Complete the square

OpenStudy (anonymous):

oh really?

OpenStudy (anonymous):

how do you know right away to do that

OpenStudy (anonymous):

Because I'm batman

OpenStudy (anonymous):

triuu

OpenStudy (anonymous):

Remember complete the square \[x^2+ax \implies x^2+2*\frac{ a }{ 2 }x + \frac{ a }{ 2 }^2 - \frac{ a }{ 2 }^2\]

OpenStudy (anonymous):

yea and with three variables?

OpenStudy (anonymous):

Yeah, do it separately for each one in one run! If that makes sense, try it out first and if you get stuck then ask for help

OpenStudy (anonymous):

coolio

OpenStudy (anonymous):

(y-3)^2-9(z+5)^2+9x^2=-441

OpenStudy (anonymous):

? @iambatman

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

then divide by -441 to get the equation to equal 1

OpenStudy (anonymous):

-441? Mhm I think you have messed up somewhere in your completing the square process

OpenStudy (anonymous):

god damn... so tedious

ganeshie8 (ganeshie8):

9x^2 + y^2 + 9z^2 - 6y - 90z + 225 = 0 as a start group the terms wid same variables together : (9x^2) + (y^2 - 6y) + (9z^2 - 90z) = -225 now complete the square for second and third groups

OpenStudy (anonymous):

Yeah, I believe it should be along the lines of \[9x^2+(y-3)^2+9(z-5)^2=9\] and then you should be able to compare it with one of the quadratic surfaces and boom boom boom, it's just this stuff that can be annoying at times :P

OpenStudy (anonymous):

awesome thank you!!!!1

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