Question Number 159941 by quvonnn last updated on 22/Nov/21
Commented by MJS_new last updated on 22/Nov/21
$$\mathrm{again}\:\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$${x}\approx\mathrm{40}.\mathrm{0163323256} \\ $$
Answered by ajfour last updated on 22/Nov/21
$${let}\:\:\:{x}={p}^{\mathrm{2}} {t}+{p}^{\mathrm{2}} {q} \\ $$$$\Rightarrow\:\frac{\mathrm{40}/{p}}{\:\sqrt{{t}+{q}−\frac{\mathrm{29}}{{p}^{\mathrm{2}} }}}+\frac{\mathrm{6}/{p}}{\:\sqrt{{t}+{q}}}=\mathrm{13} \\ $$$${let}\:\:\:\:{q}−\frac{\mathrm{29}}{{p}^{\mathrm{2}} }=−{q} \\ $$$$\Rightarrow\:\:\mathrm{2}{qp}^{\mathrm{2}} =\mathrm{29} \\ $$$$\Rightarrow\:\:\frac{\mathrm{40}}{\:\sqrt{{t}−{q}}}+\frac{\mathrm{6}}{\:\sqrt{{t}+{q}}}=\mathrm{13}{p} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1600}}{{t}−{q}}+\frac{\mathrm{36}}{{t}+{q}}=\mathrm{169}\left(\frac{\mathrm{29}}{\mathrm{2}{q}}\right)−\frac{\mathrm{480}}{\:\sqrt{{t}^{\mathrm{2}} −{q}^{\mathrm{2}} }} \\ $$$${let}\:\:\:\:\:\:\frac{\mathrm{160}×\mathrm{29}}{\mathrm{2}{q}}=\frac{\mathrm{480}}{\:\sqrt{{t}^{\mathrm{2}} −{q}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{1600}\left({t}+{q}\right)+\mathrm{36}\left({t}−{q}\right)}{{t}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\mathrm{1636}{t}+\mathrm{1564}{q}=\mathrm{0} \\ $$$$…. \\ $$