Question

Transform the equation by completing the square: x^2+y^2+6x-16y+48 is greater than or equal to 0. Explanation also please!

Answer 1

\[(x^2+6x)+(y^2-16y) \ge -48\]
\[(x^2+6x+9)+(y^2-16y+64) \ge -48+64+9\]
\[(x+3)^2+(y-8)^2 \ge 73-48\]
\[(x+3)^2+(y-8)^2 \ge 25\]

Answer 2

this means you will graph
(x+3)^2+(y-8)^2=5^2
and shade everything outside the circle

Answer 3

completeing a square can be seen as a geometric interpretation; for example:

given: x^2 + 6x + ____ , we can "complete a square" like this


\begin{array}c
&&&(6/2)x\\
&x^2&x&x&x\\
&x&&&\\
(6/2)x&x&&&\\
&x&&&\\
\end{array}

now fill in the missing parts to complete it:


\begin{array}c
&&&(6/2)x\\
&x^2&x&x&x\\
&x&1&2&3\\
(6/2)x&x&4&5&6\\
&x&7&8&9\\
\end{array}

9 missing parts soo...

x^2 + 6x +9 is a "complete square

Answer 4

oh no radar you add 9 on one side you add it to the other
just like 64

Answer 5

since the geometry can get messy, we can stick to the algebra;

ax^2 + bx + _____

we take half the x parts (that 'b') and square them to "complete the square"

ax^2 + bx + (b/2)^2

Answer 6

..r^2 = 25 :)