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^2 +(√(3−x)) = 3 x^2 = 3−(√(3−x )) x = (√(3−(√(3−x)))) x = (√(3−(√(3−(√(3−(√(3−x)) )))))) x = (√(3−(√(3−(√(3−........∞)))))) x = (√(3−x)) x^2 = 3−x x^2 +x−3=0 x= ((−1±(√(1^2 −4∙1∙(−3))))/(2∙1)) x = ((−1±(√(1+12)))/2) x = ((−1±(√(13)))/2) x = ((−1+(√(13)))/2) Answer [only positive value satisfied] [also x = −1 ]](Q170852.png)
$${solution}:\:\: \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}\:=\:\mathrm{3} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{3}−\sqrt{\mathrm{3}−{x}\:} \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−{x}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−{x}}\:}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−........\infty}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−{x}}\: \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{3}−{x} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{1}\centerdot\left(−\mathrm{3}\right)}}{\mathrm{2}\centerdot\mathrm{1}} \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{12}}}{\mathrm{2}}\: \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${x}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:{Answer}\:\:\left[{only}\:\:{positive}\:\:{value}\:\:{satisfied}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{also}\:{x}\:=\:−\mathrm{1}\:\right] \\ $$
Commented by mr W last updated on 01/Jun/22
$${nice}\:{solution}! \\ $$
Answered by MJS_new last updated on 01/Jun/22
$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{3}−{x}}={t}\geqslant\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +{t}+\mathrm{6}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$${t}\geqslant\mathrm{0}\:\Rightarrow\:{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{t}=\mathrm{2} \\ $$$$\Rightarrow\:{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{x}=−\mathrm{1} \\ $$