Mathematics 34 Online
OpenStudy (anonymous):

complete the square. 16x^2 + 16y^2 + 16x +40y - 7 =0

myininaya (myininaya):

$(16x^2+16x)+(16y^2+40y)=7$

myininaya (myininaya):

$16(x^2+x)+16(y^2+\frac{40}{16}y)=7$

myininaya (myininaya):

$16(x^2+x)+16(y^2+\frac{5}{2}y)=7$

myininaya (myininaya):

$\frac{16(x^2+x)+16(y^2+\frac{5}{2}y)}{16}=\frac{7}{16}$

myininaya (myininaya):

$(x^2+x)+(y^2+\frac{5}{2}y)=\frac{7}{16}$

myininaya (myininaya):

$(x^2+1x+(\frac{1}{2})^2)+(y^2+\frac{5}{2}y+(\frac{5}{2 \cdot 2})^2)=\frac{7}{16}+(\frac{1}{2})^2+(\frac{5}{2 \cdot 2})^2$

jimthompson5910 (jim_thompson5910):

$\large 16x^2 + 16y^2 + 16x +40y - 7 =0$ $\large 16x^2 + 16x + 16y^2 +40y - 7 =0$ $\large 16(x^2 + x) + 16(y^2 +\frac{5}{2}y) - 7 =0$ $\large 16(x^2 + x+\frac{1}{4}-\frac{1}{4}) + 16(y^2 +\frac{5}{2}y) - 7 =0$ $\large 16(x^2 + x+\frac{1}{4}-\frac{1}{4}) + 16(y^2 +\frac{5}{2}y+\frac{25}{16}-\frac{25}{16}) - 7 =0$ $\large 16((x+\frac{1}{2})^2-\frac{1}{4}) + 16((y+\frac{5}{4})^2-\frac{25}{16}) - 7 =0$ $\large 16(x+\frac{1}{2})^2+16(-\frac{1}{4}) + 16(y+\frac{5}{4})^2+16(-\frac{25}{16}) - 7 =0$ $\large 16(x+\frac{1}{2})^2-4 + 16(y+\frac{5}{4})^2-25 - 7 =0$ $\large 16(x+\frac{1}{2})^2 + 16(y+\frac{5}{4})^2-36 =0$ $\large 16(x+\frac{1}{2})^2 + 16(y+\frac{5}{4})^2=36$ $\large (x+\frac{1}{2})^2 + (y+\frac{5}{4})^2=\frac{36}{16}$ $\large (x+\frac{1}{2})^2 + (y+\frac{5}{4})^2=(\frac{6}{4})^2$ $\large (x+\frac{1}{2})^2 + (y+\frac{5}{4})^2=(\frac{3}{2})^2$ So this is an equation of a circle with center (-1/2, 5/4) and has a radius of 3/2 units.

myininaya (myininaya):

$(x^2+x+(\frac{1}{2})^2)+(y^2+\frac{5}{2}y+(\frac{5}{4})^2)=\frac{7}{16}+\frac{1}{4}+(\frac{5}{4})^2$

jimthompson5910 (jim_thompson5910):

oops meant to write So this is an equation of a circle with center (-1/2, -5/4) and has a radius of 3/2 units.

myininaya (myininaya):

$(x+\frac{1}{2})^2+(y+\frac{5}{4})^2=\frac{7}{16}+\frac{1}{4}+\frac{25}{16}$

myininaya (myininaya):

$(x+\frac{1}{2})^2+(y+\frac{5}{4})^2=\frac{7+4+25}{16}$

myininaya (myininaya):

$(x+\frac{1}{2})^2+(y+\frac{5}{4})^2=\frac{36}{16}$

myininaya (myininaya):

myininaya (myininaya):

oh you reduced it

OpenStudy (anonymous):

nothing it's just not squared

myininaya (myininaya):

lol

OpenStudy (anonymous):

i got it. but jim it's hard to follow your work. some of it just doesn't make sense

jimthompson5910 (jim_thompson5910):

$\large \frac{36}{16}=\frac{6^2}{4^2}=\left(\frac{6}{4}\right)^2=\left(\frac{3}{2}\right)^2$

myininaya (myininaya):

$(x+\frac{1}{2})^2+(y+\frac{5}{4})^2=(\frac{6}{4})^2=(\frac{3}{2})^2$

jimthompson5910 (jim_thompson5910):

which part(s)?

myininaya (myininaya):

no its good i just didn't look to see if that fraction was reducible

OpenStudy (anonymous):

like where you have 1/4 -1/4 it's not a big deal though

jimthompson5910 (jim_thompson5910):

I'm adding 0 there (since 1/4-1/4=0). I got that by taking half of the x coefficient and squaring it.

myininaya (myininaya):

instead of adding it on both sides first he subtract what he added on one side since he added 1/4 he took out 1/4 but he added 1/4 to complete the square

OpenStudy (anonymous):

o

jimthompson5910 (jim_thompson5910):

now that I look at it, it's probably a good idea to break it up a bit eh?

jimthompson5910 (jim_thompson5910):

ie not make it so long

OpenStudy (anonymous):

maybe. i think the problem i mostly had was just trying to do too much at once and it wasn't all adding up

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