(√x)+1=0 find x Mastermind


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Question Number 169396 by Mastermind last updated on 29/Apr/22

$\sqrt{x} + 1 = 0$ $f i n d x$ $M a s t e r m i n d$
Commented by mr W last updated on 29/Apr/22

$i^{4} = 1$ $\sqrt{i^{4}} + 1 = 2 \neq 0$
Commented by infinityaction last updated on 29/Apr/22

$i^{4} w h e r e i i s a c o m p l e x n u m b e r$
Commented by mr W last updated on 29/Apr/22

$\sqrt{x} + 1 = 0 h a s n o s o l u t i o n e v e n f o r$ $x \in C .$
Commented by Mastermind last updated on 29/Apr/22

$W o w!$ $t h a n k s m a n$
Commented by Mastermind last updated on 29/Apr/22

$i l o v e t h e w a y y o u a n s w e r e d i t$
Commented by Mastermind last updated on 29/Apr/22

$T h a n k y o u t o o, f o r a t t e m p t i n g i t$
Commented by Mastermind last updated on 29/Apr/22

$B u t w h y {d i d n}^{'} t y o u c o n s i d e r l a w o f$ $e x p o n e n t i a l$ $\sqrt{(x^{n})} = x^{\frac{n}{2}}$ $N o w, x = i^{4}$ $\sqrt{(i^{4})} = i^{\frac{4}{2}} \Rightarrow i^{2} = - 1$ $I w a n t y o u t o c h e c k t h i s @ m r W$
Commented by mr W last updated on 30/Apr/22

$i j u s t s a y : b e c a r e f u l! o t h e r w i s e y o u$ $w i l l g e t t h i n g s l i k e t h i s :$ $1 = {(1)}^{\frac{3}{2}} = {({(- 1)}^{2})}^{\frac{3}{2}} = {(- 1)}^{2 \times \frac{3}{2}} = {(- 1)}^{3} = - 1$
Commented by infinityaction last updated on 30/Apr/22

$i^{4} = 1$ $\sqrt{x} = - 1$ $i^{2} = - 1$ $\sqrt{x} = i^{2}$ $x = i^{4}$
Commented by mr W last updated on 30/Apr/22

$\sqrt{x} = i^{2} ⇎ x = i^{4}$ $b e c a r e f u l a g a i n!$ $w h e n y o u s q u a r e a n e q u a t i o n, y o u$ $g e t a n e w e q u a t i o n w h i c h m a y c o n t a i n$ $r o o t s w h i c h a r e n o t t h e r o o t s o f t h e$ $o r i g i n a l e q u a t i o n . s o y o u m u s t c h e c k$ $i f t h e r o o t s f u l f i l l t h e o r i g i n a l$ $e q u a t i o n . i n c u r r e n t c a s e {y o u}^{'} l l s e e$ $t h a t x = i^{4} {d o e s n}^{'} t f u l f i l l t h e o r i g i n a l$ $e q u a t i o n .$
Commented by mr W last updated on 30/Apr/22

$y o u m a y i n s i s t o n t h a t x = i^{4} i s t h e$ $r o o t o f \sqrt{x} + 1 = 0 . {t h a t}^{'} s y o u r r i g h t .$ $b u t n e v e r t h e l e s s i t i s w r o n g!$
Commented by mr W last updated on 30/Apr/22

Commented by infinityaction last updated on 30/Apr/22

$g o t i t s i r$ $t h a n k y o u s i r$
Commented by MJS_new last updated on 30/Apr/22

$steps we use to solve i . e .$ $x^{3} = - 1; x \in C$ $x^{3} = e^{i π}$ $we must search for x = e^{i θ} with e^{3 i θ} = e^{i π}$ $\Rightarrow 3 θ = π \Leftrightarrow θ = \frac{π}{3} \Rightarrow x = e^{i \frac{π}{3}} = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $but {there}^{'} s more :$ $x_{k}^{3} = e^{i (π + 2 π n)}; n = 0, 1, 2 \land k = n + 1$ $this ‘ ‘ {trick}^{″} on the rhs leads to 3 solutions .$ $we did nothing on the lhs$ $x_{1}^{3} = e^{i π} \Rightarrow x_{1} = e^{i \frac{π}{3}} = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $x_{2}^{3} = e^{3 i π} \Rightarrow x_{2} = e^{i π} = - 1$ $x_{3}^{3} = e^{5 i π} \Rightarrow x_{3} = e^{i \frac{5 π}{3}} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$ $tbese 3 solutions represent 3 different points$ $in the complex plane$ $now test the solutions$ ${(e^{i \frac{π}{3}})}^{3} = e^{i π} = - 1 true$ ${(e^{i π})}^{3} = e^{3 i π} = e^{i π} = - 1 true$ ${(e^{i \frac{5 π}{3}})}^{3} = e^{5 i π} = e^{i π} = - 1 true$ $now {let}^{'} s try$ $x^{1 / 3} = - 1; x \in C$ $x^{1 / 3} = e^{i π}$ $obviously our^{″} {trick}^{″} on the rhs {doesn}^{'} t$ $work (n \in N because we need a countable$ $number of solutions)$ $we must search for x = e^{i θ} with e^{i \frac{θ}{3}} = e^{i π}$ $\Rightarrow \frac{θ}{3} = π \Leftrightarrow θ = 3 π \Rightarrow x = e^{3 i π} = e^{i π} = - 1$ $but here we did something on the lhs$ $test : {(e^{i π})}^{1 / 3} = e^{i \frac{π}{3}} \neq - 1$ $so {here}^{'} s a totally different case$ $the rule x^{1 / 3} = - {(- x)}^{1 / 3} is only valid for x \in R$ $strictly following the rules of C x^{1 / 3} = - 1 has$ $no solution$ $now try$ $x^{1 / 2} = - 1; x \in C$ $x^{1 / 2} = e^{i π}$ $x = e^{i θ} with e^{i \frac{θ}{2}} = e^{i π}$ $\Rightarrow \frac{θ}{2} = π \Leftrightarrow θ = 2 π \Rightarrow x = e^{2 i π} = e^{0 i π} = e^{0} = 1$ $again we did something on the lhs$ $test : {(e^{0})}^{1 / 2} = e^{0} = 1 \neq - 1$ $again no solution in C and obviously none$ $in R (because \sqrt{1} = 1; also f (x) = \sqrt{x} is only$ $a half parabola)$ $\Rightarrow we need a different approach for solving$ $equations in C$
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