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Question Number 170844 by mr W last updated on 01/Jun/22
solve x^2 +(√(3−x))=3
$${solve}\:{x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$
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 ]
$${solution}:\:\: \\ $$$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}\:=\:\mathrm{3} \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{3}−\sqrt{\mathrm{3}−{x}\:} \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−{x}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−{x}}\:}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−\sqrt{\mathrm{3}−……..\infty}}}\: \\ $$$${x}\:=\:\sqrt{\mathrm{3}−{x}}\: \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{3}−{x} \\ $$$${x}^{\mathrm{2}} +{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{1}\centerdot\left(−\mathrm{3}\right)}}{\mathrm{2}\centerdot\mathrm{1}} \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{12}}}{\mathrm{2}}\: \\ $$$${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$${x}\:=\:\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:{Answer}\:\:\left[{only}\:\:{positive}\:\:{value}\:\:{satisfied}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{also}\:{x}\:=\:−\mathrm{1}\:\right] \\ $$
Commented by mr W last updated on 01/Jun/22
nice solution!
$${nice}\:{solution}! \\ $$
Answered by MJS_new last updated on 01/Jun/22
x^2 +(√(3−x))=3  (√(3−x))=t≥0  t^4 −6t^2 +t+6=0  (t−2)(t+1)(t^2 +t−3)=0  t≥0 ⇒ t=((−1+(√(13)))/2)∨t=2  ⇒ x=((−1+(√(13)))/2)∨x=−1
$${x}^{\mathrm{2}} +\sqrt{\mathrm{3}−{x}}=\mathrm{3} \\ $$$$\sqrt{\mathrm{3}−{x}}={t}\geqslant\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +{t}+\mathrm{6}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +{t}−\mathrm{3}\right)=\mathrm{0} \\ $$$${t}\geqslant\mathrm{0}\:\Rightarrow\:{t}=\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{t}=\mathrm{2} \\ $$$$\Rightarrow\:{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{13}}}{\mathrm{2}}\vee{x}=−\mathrm{1} \\ $$

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