solve-x-2-3-x-3-

Question Number 170844 by mr W last updated on 01/Jun/22

$${solve}\:{x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$

Answered by balirampatel last updated on 01/Jun/22

$${solution}:\:\: \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}\:=\:\mathrm{3} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{3}−\sqrt{\mathrm{3}−{x}\:} \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−{x}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−{x}}\:}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−……..\infty}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−{x}}\: \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{3}−{x} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{1}\centerdot\left(−\mathrm{3}\right)}}{\mathrm{2}\centerdot\mathrm{1}} \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{12}}}{\mathrm{2}}\: \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${x}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:{Answer}\:\:\left[{only}\:\:{positive}\:\:{value}\:\:{satisfied}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{also}\:{x}\:=\:−\mathrm{1}\:\right] \\ $$

Commented by mr W last updated on 01/Jun/22

$${nice}\:{solution}! \\ $$

Answered by MJS_new last updated on 01/Jun/22

$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{3}−{x}}={t}\geqslant\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +{t}+\mathrm{6}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$${t}\geqslant\mathrm{0}\:\Rightarrow\:{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{t}=\mathrm{2} \\ $$$$\Rightarrow\:{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{x}=−\mathrm{1} \\ $$

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