Question Number 98073 by bobhans last updated on 11/Jun/20
$$\mathrm{solve}\:\sqrt{\mathrm{6}−\mathrm{x}}\:=\:\mathrm{6}−\mathrm{x}^{\mathrm{2}} \\ $$
Commented by john santu last updated on 11/Jun/20
$$\mathrm{equation}\:\mathrm{defined}\:\mathrm{in}\:\mathrm{6}−\mathrm{x}\geqslant\mathrm{0}\:\wedge\:\mathrm{6}−\mathrm{x}^{\mathrm{2}} \:\geqslant\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{x}\:\leqslant\:\mathrm{6}\:\wedge\:−\sqrt{\mathrm{6}}\:\leqslant\:\mathrm{x}\:\leqslant\:\sqrt{\mathrm{6}} \\ $$$$\mathrm{squaring}\:\mathrm{6}−\mathrm{x}\:=\:\mathrm{x}^{\mathrm{4}} −\mathrm{12x}^{\mathrm{2}} +\mathrm{36} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{12x}^{\mathrm{2}} +\mathrm{x}+\mathrm{30}\:=\:\mathrm{0}\: \\ $$$$\mathrm{Horner}\:\mathrm{method} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{5}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:−\mathrm{3}\:\left(\mathrm{rejected}\right) \\ $$$$\mathrm{x}\:=\:\mathrm{2}\:\left(\mathrm{solution}\right) \\ $$$$\mathrm{x}\:=\:\frac{\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}}\:\Rightarrow\begin{cases}{\mathrm{x}=\frac{\mathrm{1}+\sqrt{\mathrm{21}}}{\mathrm{2}}\:\left(\mathrm{rejected}\right)}\\{\mathrm{x}=\frac{\mathrm{1}−\sqrt{\mathrm{21}}}{\mathrm{2}}\:\left(\mathrm{solution}\right)}\end{cases} \\ $$
Commented by bobhans last updated on 11/Jun/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Answered by MJS last updated on 11/Jun/20
![6−x=36−12x^2 +x^4 [squaring leads to false solutions] x^4 −12x^2 +x+30=0 (x−2)(x+3)(x^2 −x−5)=0 x_1 =2 x_2 =−3 x_(3, 4) =((1±(√(21)))/2) inserting in original equation gives x=2∨x=((1−(√(21)))/2)](https://www.tinkutara.com/question/Q98080.png)
$$\mathrm{6}−{x}=\mathrm{36}−\mathrm{12}{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \:\:\:\:\:\left[\mathrm{squaring}\:\mathrm{leads}\:\mathrm{to}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{false}\:\mathrm{solutions}\right] \\ $$$${x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{2}} +{x}+\mathrm{30}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{2} \\ $$$${x}_{\mathrm{2}} =−\mathrm{3} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:\mathrm{original}\:\mathrm{equation}\:\mathrm{gives} \\ $$$${x}=\mathrm{2}\vee{x}=\frac{\mathrm{1}−\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$