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Question Number 115986 by bemath last updated on 30/Sep/20

Answered by MJS_new last updated on 30/Sep/20

(√(5−x))=5−x^2   5−x≥0 ⇒ x≤5  5−x^2 ≥0 ⇒ −(√5)≤x≤(√5)  ⇒ −(√5)≤x≤(√5)  squaring both sides  5−x=(5−x^2 )^2   x^4 −10x^2 +x+20=0  (x^2 −x−4)(x^2 +x−5)=0  x_1 =((1−(√(17)))/2)  x_2 =((1+(√(17)))/2)>(√5)  x_3 =((−1−(√(21)))/2)<−(√5)  x_4 =((−1+(√(21)))/2)  ⇒  x=((1−(√(17)))/2)∨x=((−1+(√(21)))/2)

$$\sqrt{\mathrm{5}−{x}}=\mathrm{5}−{x}^{\mathrm{2}} \\ $$$$\mathrm{5}−{x}\geqslant\mathrm{0}\:\Rightarrow\:{x}\leqslant\mathrm{5} \\ $$$$\mathrm{5}−{x}^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\:−\sqrt{\mathrm{5}}\leqslant{x}\leqslant\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\:−\sqrt{\mathrm{5}}\leqslant{x}\leqslant\sqrt{\mathrm{5}} \\ $$$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{5}−{x}=\left(\mathrm{5}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +{x}+\mathrm{20}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{1}−\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{2}}>\sqrt{\mathrm{5}} \\ $$$${x}_{\mathrm{3}} =\frac{−\mathrm{1}−\sqrt{\mathrm{21}}}{\mathrm{2}}<−\sqrt{\mathrm{5}} \\ $$$${x}_{\mathrm{4}} =\frac{−\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}=\frac{\mathrm{1}−\sqrt{\mathrm{17}}}{\mathrm{2}}\vee{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$

Commented by bemath last updated on 30/Sep/20

thank you prof

$${thank}\:{you}\:{prof} \\ $$

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