Question Number 106906 by Study last updated on 07/Aug/20
Answered by JDamian last updated on 07/Aug/20
$$\mathrm{3} \\ $$
Answered by Dwaipayan Shikari last updated on 07/Aug/20
$$\sqrt{\mathrm{20}+\sqrt{\mathrm{20}+\sqrt{\mathrm{20}}}…_{} }={a} \\ $$$$\mathrm{20}+{a}={a}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −{a}−\mathrm{20}=\mathrm{0} \\ $$$${a}=\mathrm{5} \\ $$$$\sqrt{\mathrm{12}+\sqrt{\mathrm{12}+\sqrt{\mathrm{12}}}.}={p} \\ $$$$\mathrm{12}+{p}={p}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} −{p}−\mathrm{12}=\mathrm{0} \\ $$$${p}=\mathrm{4} \\ $$$$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}={t} \\ $$$$\mathrm{6}+{t}={t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −{t}−\mathrm{6}=\mathrm{0} \\ $$$${t}=\mathrm{3} \\ $$$$\frac{{a}+{p}}{{t}}=\mathrm{3} \\ $$