Question 1115399: Find the values for A, B and C if y^2 - 6x +16y + 94 = (y+C)^2 -B(x+A)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! start with y^2 - 6x + 16y + 94 = (y+C)^2 -B(x+A)
the left side of your equation is y^2 - 6x + 16y + 94.
rearrange and group the y terms to get (y^2 + 16y) - 6x + 94.
(y^2 + 16y) is equal to (y + 8) ^ 2 - 64.
this is because (y + 8) ^ 2 is equal to y^2 + 16y + 64, and, if you subtract 64 from both sides of that equation, you get (y + 8) ^ 2 - 64 = y^2 + 16y.
your expression on the left side of the equation becomes (y + 8) ^ 2 - 64 - 6x + 94.
combine like terms to get (y + 8) ^ 2 - 6x + 30
group the x and constant terms together and factor out the 6 to get:
(y + 8) ^ 2 - 6 * (x - 5).
your expression on the right side of the original equation is (y + C) ^ 2 - B * (x + A).
your equation becomes (y + 8) ^ 2 - 6 ( (x - 5) = (y + C) ^ 2 - B * (x + A).
this results in:
C = 8
B = 6
A = -5
when you replace A, B, and C with their respective values, the right side of your equation becomes (y + 8) ^ 2 - 6 * (x - 5), which is identical to the expression on the left side of your equation.
that means you're done.
the method you used was the completing the square method.
a reference on that method can be found at http://www.purplemath.com/modules/solvquad3.htm
any questions, give me a shout.
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