Question Number 135910 by liberty last updated on 17/Mar/21
$${Algebra}\: \\ $$$$\:\:\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:{p}+{q}\sqrt{\mathrm{2}} \\ $$$${then}\:\frac{{q}^{\mathrm{2}} }{{p}\sqrt{\mathrm{2}}}\:+\:{p}^{\mathrm{2}} {q}\:=? \\ $$
Answered by som(math1967) last updated on 17/Mar/21
$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{4}+\mathrm{2}−\mathrm{2}.\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}×\mathrm{2}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{9}+\mathrm{2}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{2}}={p}\:+{q}\sqrt{\mathrm{2}} \\ $$$$\therefore{p}=\mathrm{3}\:,\:\:{q}=\mathrm{1} \\ $$$$\:\frac{{q}^{\mathrm{2}} }{{p}\sqrt{\mathrm{2}}}\:+{p}^{\mathrm{2}} {q}=\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\:+\mathrm{9} \\ $$