Question

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

Answer 1

\[(16x^2+16x)+(16y^2+40y)=7\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

\[\large 16x^2 + 16y^2 + 16x +40y - 7 =0\]


\[\large 16x^2 + 16x + 16y^2 +40y - 7 =0\]


\[\large 16(x^2 + x) + 16(y^2 +\frac{5}{2}y) - 7 =0\]


\[\large 16(x^2 + x+\frac{1}{4}-\frac{1}{4}) + 16(y^2 +\frac{5}{2}y) - 7 =0\]


\[\large 16(x^2 + x+\frac{1}{4}-\frac{1}{4}) + 16(y^2 +\frac{5}{2}y+\frac{25}{16}-\frac{25}{16}) - 7 =0\]


\[\large 16((x+\frac{1}{2})^2-\frac{1}{4}) + 16((y+\frac{5}{4})^2-\frac{25}{16}) - 7 =0\]


\[\large 16(x+\frac{1}{2})^2+16(-\frac{1}{4}) + 16(y+\frac{5}{4})^2+16(-\frac{25}{16}) - 7 =0\]


\[\large 16(x+\frac{1}{2})^2-4 + 16(y+\frac{5}{4})^2-25 - 7 =0\]


\[\large 16(x+\frac{1}{2})^2 + 16(y+\frac{5}{4})^2-36 =0\]


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


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


\[\large (x+\frac{1}{2})^2 + (y+\frac{5}{4})^2=(\frac{6}{4})^2\]


\[\large (x+\frac{1}{2})^2 + (y+\frac{5}{4})^2=(\frac{3}{2})^2\]


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

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

what happen to your radius?

Answer 14

oh you reduced it

Answer 15

nothing it's just not squared

Answer 16

lol

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

which part(s)?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

o

Answer 26

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

Answer 27

ie not make it so long

Answer 28

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