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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
 2 years ago
 2 years 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
 2 years ago
 2 years 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.
 2 years ago

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