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nooo here the autor neglected the - ve root from the square root thise equation has 2 possible solutions(3 - x ) = + or - ( bi - x )here he neglected negative root
You were wrong in the second step, why would you times it by (pi-3) ? The final result just shown your mistake, that means in the second step, you times each side with zero as (3#pi-3) = 0 and this is wrong since you can not times or divide anything with zero.
It is wrong. If you start backwards. Ie PI = 3 and make operations backwards, you have to divide by PI - 3 (3rd step up and down), but PI - 3 = 0, then this operation is prohibited.
The wrong step is how you get 2nd equation from the bottom from 3rd step from the bottom. The right next equation after (3-x)^2=(PI-x)^2 is this equation => |3-x|=|PI-x|
I can prove that very easy... x=(PI+3)/2 and when you put that in |3-x|=|PI-x| you have right equation => (~ -0.14)^2 = (~ +0.14)^2 but you cant say that ~ -0.14 = ~ +0.14 ;)
Please prove it if it is wrong
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The solution steps are correct if we consider that (pi) is a mathematical symbol just like y,z,....etc
The only right answer!
This is WRONG !! A lot of mistakes here but the most important is that, you should make up your mind whether to use Pi as a variable or constant, because you used it as a constant in the second step when you where planing to use it as a variable.
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