Question

solve by completing the square: x^2+11x-1=0

Answer 1

\[x^2+11x=1\]
\[(x+\frac{11}{2})^2=1+\frac{121}{4}=\frac{125}{4}\]
\[x+\frac{11}{2}=\pm\frac{\sqrt{125}}{2}\]
\[x+\frac{11}{2}=\pm\frac{5\sqrt{5}}{2}\]
\[x=-\frac{11\pm5\sqrt{5}}{2}\]

Answer 2

sorry last step should be
\[x=\frac{-11\pm5\sqrt{5}}{2}\] i had the minus sign in the wrong place

Answer 3

x^2+11x-1=0 tric to complete square
(1) divide by the coefficient of x^2
x^2+11x-1=0
(2) move the constant to the other side
x^2+11x=1
(3) tahe half of the coefficient of x,square it and add it to both side
x^2+11x+(11/2)^2=1+(11/2)^2
x^2+11x+(11/2)^2=125/4
x^2+2.11/2X+(11/2)^2=125/4
(x+11/2)^2=125/4
sqrting both side
x+11/2=5sqrt5/2
therefore x=[-11(+-) *5 sqrt5]/2