Question-192936

Question Number 192936 by Abdullahrussell last updated on 31/May/23

Answered by Frix last updated on 31/May/23

Obviously x^{2} = 3 \lor x^{2} = 4 \Rightarrow

x = \pm \sqrt{3} \lor x = \pm 2

If we use \sqrt[3]{- z} = - \sqrt[3]{z} also x^{2} = 12 \Rightarrow x = \pm 2 \sqrt{3}

Answered by MM42 last updated on 31/May/23

\sqrt{x^{2} - 3} = u \Rightarrow x^{2} = u^{2} + 3

\Rightarrow u + \sqrt[3]{1 - u^{2}} = 1 \Rightarrow 1 - u^{2} = {(1 - u)}^{3}

\Rightarrow u (u - 1) (u - 3) = 0

\Rightarrow u = 0 \Rightarrow x^{2} = 3 \Rightarrow x = \pm \sqrt{3}

& u = 1 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2

& u = 3 \Rightarrow x^{2} = 12 \Rightarrow x = \pm 2 \sqrt{3}

Commented by MM42 last updated on 31/May/23

t h e d i a g r a m b e l o w s h o w s t h e s t a t e [

o f t h e t h r e a d s

f (x) = \sqrt{x^{2} - 3} + \sqrt[3]{4 - x^{2}} - 1

Commented by MM42 last updated on 31/May/23

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