Question

thank

Answer 1

Hey girl :)
We need to complete the square on the x's and on the y's separately.\[\large\rm \color{orangered}{5x^2+65x}+\color{royalblue}{5y^2-15y}-75=0\]So let's group our x's together like this, in orange,
and our y's together, in blue.

Answer 2

Factoring a 5 out of each x term gives us\[\large\rm \color{orangered}{5(x^2+13x)}+\color{royalblue}{5y^2-15y}-75=0\](Since 65 divided by 5 is 13, that gives us a 13 on the x.)

Answer 3

Do you remember how to `complete the square`? :O

Answer 4

not really;/

Answer 5

The 13 is our b coefficient.
We cut it in half,\[\large\rm \frac{13}{2}\]Then we square it,\[\large\rm \frac{169}{4}\]This is the the value that completes the square for us,
so we'll add it to each side,\[\large\rm \color{orangered}{5\left(x^2+13x+\frac{169}{4}\right)}+\color{royalblue}{5y^2-15y}-75=5\cdot\frac{169}{4}\]But on the right side we'll have to multiply it by 5,
since it's being multiplied by 5 on the left side.
And we now have a perfect square trinomial.
It will factor down to \(\large\rm \left(x+\frac{b}{2}\right)^2\)
So if we go back to our 13/2, that tells us that our perfect square trinomial will factor into\[\large\rm \color{orangered}{5\left(x+\frac{13}{2}\right)^2}+\color{royalblue}{5y^2-15y}-75=5\cdot\frac{169}{4}\]So that finishes the standard form on the x's.

Answer 6

These numbers are really awful...
why didn't they give us nice numbers to work with? UGHH

Answer 7

Oh I guess I'm now realizing.. everything is divisible by 5,
let's take a 5 out of EVERYTHING.\[\large\rm \color{orangered}{\left(x+\frac{13}{2}\right)^2}+\color{royalblue}{y^2-3y}-15=\frac{169}{4}\]Ok that's a little easier to deal with.

Answer 8

Try to do the same with the y's :U

Answer 9

^2-3?

Answer 10

3^2y @zepdrix

Answer 11

? 0_o