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jhonyy9
how is it possible ? x=(pi +3)/2 2x=pi +3 2x(pi -3)=(pi +3)(pi -3) 2pi*x -6x =(pi)^2 -9 9 -6x =(pi)^2 -2pi*x 9 -6x +x^2 =(pi)^2 -2pi*x +x^2 (3 -x)^2 =(pi -x)^2 3-x = pi -x pi =3
Everything looks legal, maybe when taking sq roots there is something that gets twisted.
If you take the negative root, u get the start equation.....
look at the 3rd line if pi = 3 then both sides of equation = 0
Didn't consider the plus or minus when taking the square root. We get \[3-x= \pm(\pi-x) \implies 3-x=\pi - x~or~3-x=x-\pi\] The first equation generates the contradiction from the problem. The second generates the value of x given at the get-go.