Completing the Squares
F(x) =x^2+3x+2
F(x)=0
X^2+3x+2=0
WHY DIDN'T IT BECOME -X^2-3X-2 ? ( EXPLAIN THIS PLEASE)
Next for the solution
X^2+3x+3(3/2)^2 = -2+(3/2)^2
(X+3/2)^2=-2+9/4
(X+3/2)^2 = 9/4 -8/4
=1/4
( WHERE DID THEY GET 8/ 4?
We want to solve x^2+3x+2=0. We already have 0 alone so we don't need to subtract anything, but of course we can subtract LHS from RHS and get -x^2-3x-2=0 but that is the same as just multiplying both sides in the equation with -1 which is of no help.
The idea of completing the square is to rewrite to equation in the form (x+something)^2=something which is easy to solve. Now (x+3/2)^2=x^2+3x+9/4 which is almost our expression. Subtracting 1/4 we get (x+3/2)^2-1/4=x^2+3x+9/4-1/4=x^3+3x+2 and we have completed the square. In your textbook the write x^2+3x=-2 (subtracting -2 on both sides) and (x+3/2)^2=-2+9/4 and since -2=-8/4 we get (x+3/29^2=9/4-8/4=1/4.
This equation can in fact been solved more easily with factorization x^2+3x+2=(x+1)(x+2) so the solutions are -1 and -2. All equations of the form x^2+(a+b)x+ab=0 has the solutions -a and -b.