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Question Number 174269 by mr W last updated on 28/Jul/22

$$solve{(x^2-3)}^2=x+3$$

Answered by Frix last updated on 28/Jul/22

$$\mathrm{obviously}x=1\vee x=-2\mathrm{the}\mathrm{other}2\mathrm{are}\mathrm{easy}$$

Answered by mr W last updated on 28/Jul/22

$${(x^2-3)}^2-3=x$$$$let{x}^2-3=t...\left(i\right)$$$$\Rightarrow t^2-3=x...\left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$x^2-t^2=t-x$$$$(t-x)(t+x+1)=0$$$$\Rightarrow t-x=0$$$$\Rightarrow x^2-x-3=0$$$$\Rightarrow x=\frac{1\pm \sqrt{13}}{2}$$$$\Rightarrow t+x+1=0$$$$\Rightarrow x^2+x-2=0$$$$\Rightarrow x=-2or1$$

Commented by Tawa11 last updated on 28/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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