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jhonyy9
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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
3x = pi x
pi =3
 one year ago
 one year ago
jhonyy9 Group Title
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 3x = pi x pi =3
 one year ago
 one year ago

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radar Group TitleBest ResponseYou've already chosen the best response.0
Everything looks legal, maybe when taking sq roots there is something that gets twisted.
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
If you take the negative root, u get the start equation.....
 one year ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
look at the 3rd line if pi = 3 then both sides of equation = 0
 one year ago

AnimalAin Group TitleBest ResponseYou've already chosen the best response.0
Didn't consider the plus or minus when taking the square root. We get \[3x= \pm(\pix) \implies 3x=\pi  x~or~3x=x\pi\] The first equation generates the contradiction from the problem. The second generates the value of x given at the getgo.
 one year ago
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