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solve-x-2-5x-6-x-2-5x-4-x-2-x-2-5x-6-x-2-5x-4-x-2-2-x-2-4-find-x-

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Question Number 171759 by Mikenice last updated on 20/Jun/22
solve: ((√(x^2 −5x+6)) + (√(x^2 −5x+4)) )^(x/2) + ((√(x^2 −5x+6)) − (√(x^2 −5x+4)))^(x/2) = 2^((x+2)/4) . find x
$${solve}: \\ $$$$\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:+\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\:\right)^{\frac{{x}}{\mathrm{2}}} \:+\:\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:−\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\right)^{\frac{{x}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{{x}+\mathrm{2}}{\mathrm{4}}} .\:{find}\:{x} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jun/22
((√(x^2 −5x+6)) + (√(x^2 −5x+4)) )^(x/2) + ((√(x^2 −5x+6)) − (√(x^2 −5x+4)))^(x/2) = 2^((x+2)/4) . find x Let (x/2)=y⇒x=2y ((√(4y^2 −10y+6)) +(√(4y^2 −10y+4)) )^y +((√(4y^2 −10y+6)) −(√(4y^2 −10y+4)) )^y =2^((y+1)/2) •a=((√(4y^2 −10y+6)) +(√(4y^2 −10y+4)) )^y (1/a)=((((√(4y^2 −10y+6)) −(√(4y^2 −10y+4)) )^y )/(((√(4y^2 −10y+6)) +(√(4y^2 −10y+4)) )^y ((√(4y^2 −10y+6)) −(√(4y^2 −10y+4)) )^y )) (1/a)=((((√(4y^2 −10y+6)) −(√(4y^2 −10y+4)) )^y )/(( (4y^2 −10y+6) −(4y^2 −10y+4) )^y )) (1/a)=((((√(4y^2 −10y+6)) −(√(4y^2 −10y+4)) )^y )/(2^y )) •((√(4y^2 −10y+6)) −(√(4y^2 −10y+4)) )^y =(2^y /a) a+(2^y /a)=2^((y+1)/2) .... ...
$$\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:+\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\:\right)^{\frac{{x}}{\mathrm{2}}} \:+\:\left(\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\:−\:\sqrt{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{4}}\right)^{\frac{{x}}{\mathrm{2}}} \:=\:\mathrm{2}^{\frac{{x}+\mathrm{2}}{\mathrm{4}}} .\:{find}\:{x} \\ $$$${Let}\:\frac{{x}}{\mathrm{2}}={y}\Rightarrow{x}=\mathrm{2}{y} \\ $$$$\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:+\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} +\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} =\mathrm{2}^{\frac{{y}+\mathrm{1}}{\mathrm{2}}} \\ $$$$\bullet{a}=\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:+\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} \\ $$$$\frac{\mathrm{1}}{{a}}=\frac{\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} }{\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:+\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} \left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} } \\ $$$$\frac{\mathrm{1}}{{a}}=\frac{\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} }{\left(\:\left(\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}\right)\:−\left(\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}\right)\:\right)^{{y}} } \\ $$$$\frac{\mathrm{1}}{{a}}=\frac{\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} }{\mathrm{2}^{{y}} \:} \\ $$$$\bullet\left(\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{6}}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{10}{y}+\mathrm{4}}\:\right)^{{y}} =\frac{\mathrm{2}^{{y}} }{{a}} \\ $$$$\:\:\:\:{a}+\frac{\mathrm{2}^{{y}} }{{a}}=\mathrm{2}^{\frac{{y}+\mathrm{1}}{\mathrm{2}}} \\ $$$$…. \\ $$$$… \\ $$
Commented by Mikenice last updated on 21/Jun/22
please sir, help me complete the solution sir
$${please}\:{sir},\:{help}\:{me}\:{complete}\:{the}\:{solution}\:{sir} \\ $$

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