Question

Rewrite x^2-2x+8y + 25=0 in the form (x-h)^2=4p(y-k)

Answer 1

Please help!

Answer 2

\[x^2-2x+8y + 25=0 \]\[\implies x^2-2x=-8y - 25\]\[\implies x^2-2x+1=-8y - 25+1\]\[\implies \left( x-1 \right)^2=-8y - 24\]\[\implies \left( x-1 \right)^2=-8(y +3)\]

Answer 3

\[\huge\implies \left( x-1 \right)^2=4 \cdot (-2)\left[y-\left( -3 \right)\right]\]

Answer 4

Okay thank you. How can I decide what to do? I mean I never know what steps to take?

Answer 5

First move all the x terms on one side & all the non-x terms on the other...

Then you need to add to both sides a number that would make the x side a perfect square.
Generally if you have x^2+ bx you would add: \[\left( \frac b2 \right)^2 or\ {b^2 \over 4}\]