Question Number 167489 by mathlove last updated on 18/Mar/22
$${a}+\sqrt{{x}}=\mathrm{2} \\ $$$${b}+\sqrt[{\mathrm{4}}]{{x}}=\mathrm{9}\:\:\:\:\:\:\:\:\:\overset{\:\:\:\:\:\:\left(\mathrm{4}{a}−{b}^{\mathrm{2}} \right)=?} {\:} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Mar/22
$${a}+\sqrt{{x}}=\mathrm{2}…\left({i}\right) \\ $$$${b}+\sqrt[{\mathrm{4}}]{{x}}=\mathrm{9}…\left({ii}\right)\:\:\:\:\:\:\:\:\overset{\:\:\:\:\:\:\left(\mathrm{4}{a}−{b}^{\mathrm{2}} \right)=?} {\:} \\ $$$$\left({i}\right)\Rightarrow\sqrt{{x}}\:=\mathrm{2}−{a}……….\left({iii}\right) \\ $$$$\left({ii}\right)\Rightarrow\sqrt[{\mathrm{4}}]{{x}}\:=\mathrm{9}−{b}\Rightarrow\sqrt{{x}}\:=\left(\mathrm{9}−{b}\right)^{\mathrm{2}} …\left({iv}\right) \\ $$$$\left({iii}\right)\:\&\:\left({iv}\right):\mathrm{2}−{a}=\left(\mathrm{9}−{b}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}−{a}=\mathrm{81}+{b}^{\mathrm{2}} −\mathrm{18}{b} \\ $$$${a}=−{b}^{\mathrm{2}} +\mathrm{18}{b}−\mathrm{79} \\ $$$$\mathrm{4}{a}−{b}^{\mathrm{2}} =\mathrm{4}\left(−{b}^{\mathrm{2}} +\mathrm{18}{b}−\mathrm{79}\right)−{b}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{5}{b}^{\mathrm{2}} +\mathrm{72}{b}−\mathrm{316} \\ $$$${b}^{\mathrm{2}} −\mathrm{18}{b}+{a}+\mathrm{79}=\mathrm{0} \\ $$$${b}=\frac{\mathrm{18}\pm\sqrt{\mathrm{18}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{316}}}{\mathrm{2}}\:=\frac{\mathrm{18}\pm\mathrm{2}\sqrt{\mathrm{2}−{a}}}{\mathrm{2}} \\ $$$$\:\:=\frac{\mathrm{18}\pm\mathrm{2}\sqrt{\left(\mathrm{9}−{b}\right)^{\mathrm{2}} }}{\mathrm{2}}=\mathrm{9}\pm\left(\mathrm{9}−{b}\right)=\mathrm{9}\pm\mathrm{9}\mp{b} \\ $$$${b}\pm{b}=\pm\mathrm{18}\Rightarrow{b}=\mathrm{9} \\ $$$${b}=\mathrm{9}\Rightarrow{a}=−{b}^{\mathrm{2}} +\mathrm{18}{b}−\mathrm{79}=−\mathrm{9}^{\mathrm{2}} +\mathrm{18}\centerdot\mathrm{9}−\mathrm{79} \\ $$$$\:\:\:\:\:\:\:\:=−\mathrm{160}+\mathrm{162}=\mathrm{2} \\ $$$$\mathrm{4}{a}−{b}^{\mathrm{2}} =\mathrm{4}\left(\mathrm{2}\right)−\mathrm{9}^{\mathrm{2}} =\mathrm{8}−\mathrm{81}=−\mathrm{73} \\ $$
Commented by mathlove last updated on 18/Mar/22
$${thanks}\:{bro} \\ $$