Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit. x2 + 2x + y2 + 4y = 19
@agent0smith
@mjdennis @.Sam. last question then finals :)
If you're back, take a look at this and I'll explain it further
one second let me read it
okay i read it
I'm stuck on that complex step
\[(x^2 + 2x+?) + (y^2 + 4y+?) = 19 \] So, the half of 2 is 1, the squared of it is 1, as for 'y' the half of it is 2, squared it you get 4, add these to your RHS as well \[(x^2 + 2x+1) + (y^2 + 4y+4) = 19 + 1+4 \] \[(x^2 + 2x+1) + (y^2 + 4y+4) =24 \]
what what
wait what*
lol, right, from here you'll have to find the question mark and you'll take the 'Number' from 2x, and 4y. then, there's a method you'll need to make, take the half of 2x and 4y and square it, but without the x and y. So, for x, half of 2 is 1, square of it is still 1. For you, half of 4 is 2, take the square of it is 4 again. Then you'd put 1 and 4 accordingly \[(x^2 + 2x+1) + (y^2 + 4y+4) = 19 + 1+4\]
ok where does the 1 and 4 go in the parenthesis?
I mean For 'y', not 'For you'
Put them as a number inside the parenthesis, with \((x^2+2x)\)
You'd get \[(x^2 + 2x+1) + (y^2 + 4y+4) = 19 + 1+4\]
yes but after they're in there where do they go?
Let them be there because you'll need to factorize the two equations later
okay, I have that written down, now what?
Great now factorize x^2+2x+1
what do you mean by that?
Factorize into this I mean \[(x+1)^2\]
And you'll need to factorize y2+4y+4 as well
so \[(y+4)^2\]
?
Nope your values are off, if you expand it you won't get y2+4y+4
Try (y+2)^2
oh bc 2 is the square root of 4?
then I would square root the whole left side right? so it would look like \[(x+1) + (y + 2) = 24\]
No just normal factorizing equation|dw:1464796298099:dw|
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