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x^2-4x-8y-28=0

in solving this kind of problem, first, group the equation like this...

(x^2-4x) - (8y+28) = 0

then, complete the equation like this...

(x^2-4x+4) - (8y+28+4) = 0

how I get this answer? simple...

[1/2(b)]^2 = [1/2(4)]^2 = 4.

after, if you get 4... add 4 to the constant to balance..

then, factor the equation..

(x-2)^2 - (8y+32) = 0

(x-2)^2 - 8(y+4) = 0

that's the final answer...