Question Number 1181 by 22 last updated on 11/Jul/15
$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}=? \\ $$
Answered by Rasheed Ahmad last updated on 24/Jul/15
$${In}\:{certain}\:{cases}\:\sqrt{{a}+{b}\sqrt{{c}}\:}\:{can}\:{be} \\ $$$${simplified}\:{into}\:{p}+{q}\sqrt{{c}}\:{form}\left({not}\right. \\ $$$$\left.{in}\:{all}\:{cases}\right).\:{The}\:{procedure}\:{is}\:{as} \\ $$$${follows} \\ $$$${Let}\:\sqrt{{a}+{b}\sqrt{{c}}}\:={p}+{q}\sqrt{{c}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}+{b}\sqrt{{c}}\:=\left({p}+{q}\sqrt{{c}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={p}^{\mathrm{2}} +{q}^{\mathrm{2}} {c}+\mathrm{2}{pq}\sqrt{{c}}\: \\ $$$${By}\:{comparing}\:{the}\:{the}\:{coefficientsss} \\ $$$${of}\:\sqrt{{c}}\:\:{and}\:{the}\:{terms}\:{not}\:{containing} \\ $$$$\sqrt{{c}\:}\:{we}\:{have}\: \\ $$$${a}={p}^{\mathrm{2}} +{q}^{\mathrm{2}} {c}\:\:{and}\:{b}=\mathrm{2}{pq} \\ $$$${If}\:{these}\:{two}\:{simultaneous}\:{eqns} \\ $$$${have}\:{solution}\:{for}\:{rational}\:{p}\:{and} \\ $$$${q}\:\:{the}\:{given}\:{expression}\:{is} \\ $$$${transformable}\:{into}\:{the}\:{form} \\ $$$${p}+{q}\sqrt{{c}}\:\:\:{otherwise}\:{not}.\: \\ $$$$\:\:{Now}\:{in}\:{this}\:{particular}\:{situation} \\ $$$$\:\:\:\:\:{a}=\mathrm{2}\:\:{b}=\mathrm{1}\:{and}\:{c}=\mathrm{2} \\ $$$${Hence}\:\:{p}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{2}} =\mathrm{2}\:\:{and}\:\mathrm{2}{pq}=\mathrm{1} \\ $$$${From}\:{the}\:{latter}\:{eqn}\:{p}=\frac{\mathrm{1}}{\mathrm{2}{q}}\:\:{and}\:{by} \\ $$$${substituting}\:{this}\:{in}\:{the}\:{former} \\ $$$${eqn}\:{we}\:{get}\: \\ $$$$\:\:\:\:\:\:\mathrm{8}{q}^{\mathrm{4}} −\mathrm{8}{q}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{U}{nfortinuately}\:{this}\:{is}\:{not} \\ $$$${solveable}\:{for}\:{rational}\:{q}. \\ $$$${So}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\:{can}\:{be}\:{represented} \\ $$$${in}\:{only}\:{in}\:{decimal}\:{form}\:{which}\:{is} \\ $$$${approximately}\:\:\mathrm{1}.\mathrm{847759}… \\ $$