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

Answered by Frix last updated on 31/May/23

$$\mathrm{Obviously}{x}^2=3\vee x^2=4\Rightarrow$$$$x=\pm \sqrt{3}\vee x=\pm 2$$$$\mathrm{If}\mathrm{we}\mathrm{use}\sqrt[3]{-z}=-\sqrt[3]{z}\mathrm{also}{x}^2=12\Rightarrow x=\pm 2\sqrt{3}$$

Answered by MM42 last updated on 31/May/23

$$\sqrt{x^2-3}=u\Rightarrow x^2=u^2+3$$$$\Rightarrow u+\sqrt[3]{1-u^2}=1\Rightarrow 1-u^2={(1-u)}^3$$$$\Rightarrow u(u-1)(u-3)=0$$$$\Rightarrow u=0\Rightarrow x^2=3\Rightarrow x=\pm \sqrt{3}$$$$\mathrm{\&}u=1\Rightarrow x^2=4\Rightarrow x=\pm 2$$$$\mathrm{\&}u=3\Rightarrow x^2=12\Rightarrow x=\pm 2\sqrt{3}$$$$$$

Commented by MM42 last updated on 31/May/23

$$thediagrambelowshowsthestate[$$$$ofthethreads$$$$f\left(x\right)=\sqrt{x^2-3}+\sqrt[3]{4-x^2}-1$$

Commented by MM42 last updated on 31/May/23

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