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