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Question Number 99227 by Algoritm last updated on 19/Jun/20

Commented by Algoritm last updated on 19/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Commented by MJS last updated on 19/Jun/20

$$\left(\mathrm{I}\right)x\sqrt{10-x^2}=x^2-6$$$$x^2(10-x^2)={(x^2-6)}^2$$$$-2x^4+22x^2-36=0$$$$x^4-11x^2-18=0$$$$(x^2-9)(x^2-2)=0$$$$\mathrm{now}\mathrm{test}\mathrm{all}\mathrm{solutions}\mathrm{if}\mathrm{they}\mathrm{fit}\left(\mathrm{I}\right)$$$$\Rightarrow x_1=-\sqrt{2}\wedge x_2=3$$

Answered by MJS last updated on 19/Jun/20

$$-\sqrt{2}<x<3$$

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