\({-0.1x^{2} - 0.6x + 0.4} = 0\)
I grouped the variables together and factored out a common term: \({-0.1(x^2 + 6x) + 0.4} = 0\)
I complete the square within the brackets: \({-0.1(x^2 + 6x + ({6\over2})^2 + 0.4} = 0\)
\({-0.1(x^2 + 6x + 9) + 0.4 - 0.9} = 0\) -> i balanced the equation
\({-0.1(x + 3)^2 - 0.5} = 0\)
and now i cant isolate the squared term:
\({-0.1(x + 3)^2} = 0.5\)
\({(x + 3)^2} = -5\)
which isn't possible.
however, the answer in the textbook is \(x = {-3 \pm \sqrt{13}}\). what did i do wrong?
Look at the 0.9's on the third line down.
The first one, coming from when the bracket is multiplied out, will be negative, so the one at the end has to be positive.
-0.1x^2 - 0.6x + 0.4 = 0
First off....the easiest thing to do is to multiply everything through by -10....this gives us
x^2 + 6x - 4 = 0 take 1/2 of 6 = 3....square it = 9....add and subtract it
x^2 + 6x + 9 - 4 - 9 = 0 factor the first three terms and simplify the rest
(x + 3)^2 - 13 = 0 add 13 to both sides
(x + 3)^2 = 13 take both roots
x + 3 = ±√13 subtract 3 from both sides
x = ±√13 - 3 or -3 ±√13 if you prefer!!!!