Question 966683


to complete the square for {{{4x^2+25y^2-33x+50y-311=0}}} first group {{{x}}} terms together and {{{y}}} terms together 

{{{(4x^2-33x)+(25y^2+50y)-311=0}}}


{{{4(x^2-(33/4)x)+25(y^2+2y)-311=0}}}...recall the square of difference {{{(a-b)^2=a^2-2ab+b^2}}}=> so, we need to add and subtract {{{b^2}}}

{{{4(x^2-(33/4)x+b^2)-b^2+25(y^2+2y+b^2)-b^2-311=0}}}

...since for {{{x}}} variable we have {{{a=1}}} and {{{2ab=33/4}}}, we can calculate {{{b}}}

{{{2ab=33/4}}}
{{{2*1b=33/4}}}
{{{b=33/8}}}

we do same with {{{y}}} variable where {{{a=1}}} and {{{2ab=2}}}=>{{{b=1}}}

then our equation will be:

{{{4(x^2-(33/8)x+(33/8)^2)-4(33/8)^2+25(y^2+2y+1^2)-25*1^2-311=0}}}


{{{4(x^2-(33/8))^2-(1089/16)+25(y+1)^2-25-311=0}}}


{{{4(x^2-(33/8))^2-(1089/16)+25(y^2+2y+2^2)-(336*16/16)=0}}}


{{{4(x^2-(33/4))^2-(1089/16)+25(y^2+2y+2^2)-(5376/16)=0}}}


{{{4(x-33/8)^2+25(y+1)^2-6465/16=0}}}


{{{4(x-33/8)^2+25(y+1)^2=6465/16}}}