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Question-84820




Question Number 84820 by Power last updated on 16/Mar/20
Answered by MJS last updated on 16/Mar/20
super easy  (√(2x−5))=t ⇔ x=(1/2)t^2 +(5/2)∧t≥0  (√((1/2)t^2 +(5/2)−2+t))+(√((1/2)t^2 +(5/2)+2+3t))=7(√2)  (√((1/2)t^2 +t+(1/2)))+(√((1/2)t^2 +3t+(9/2)))=7(√2)  (√(((t+1)^2 )/2))+(√(((t+3)^2 )/2))=7(√2)  ((t+1)/( (√2)))+((t+3)/( (√2)))=7(√2)  2t+4=14  t=5  x=15
$$\mathrm{super}\:\mathrm{easy} \\ $$$$\sqrt{\mathrm{2}{x}−\mathrm{5}}={t}\:\Leftrightarrow\:{x}=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}\wedge{t}\geqslant\mathrm{0} \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{2}+{t}}+\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{2}}+\mathrm{2}+\mathrm{3}{t}}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +{t}+\frac{\mathrm{1}}{\mathrm{2}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{3}{t}+\frac{\mathrm{9}}{\mathrm{2}}}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}}+\sqrt{\frac{\left({t}+\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{2}}}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$$\frac{{t}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{t}+\mathrm{3}}{\:\sqrt{\mathrm{2}}}=\mathrm{7}\sqrt{\mathrm{2}} \\ $$$$\mathrm{2}{t}+\mathrm{4}=\mathrm{14} \\ $$$${t}=\mathrm{5} \\ $$$${x}=\mathrm{15} \\ $$
Commented by Power last updated on 16/Mar/20
thanks
$$\mathrm{thanks} \\ $$

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