Question

Solve x^2-x-8=0 by completing the square, correct to 2 places of decimals.

Answer 1

anybody?

Answer 2

i can do it using the formula but having problem with this method

Answer 3

\[x^2-x-8=0\]\[(x-\frac{1}{2})^2-(\frac{1}{2})^2-8=0\]\[(x-\frac{1}{2})^2=\frac{1}{4}+8\]\[(x-\frac{1}{2})^2=\frac{33}{4}\]\[x-\frac{1}{2}=\pm \sqrt{\frac{33}{4}}\]\[x=\frac{1}{2} \pm \sqrt{\frac{33}{4}}\]

Answer 4

aahhh - i get it - thanx very much

Answer 5

;)