Find-the-square-root-of-7-5-

Question Number 91047 by I want to learn more last updated on 27/Apr/20

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{square}\:\mathrm{root}\:\mathrm{of}:\:\:\:\:\sqrt{\mathrm{7}}\:\:+\:\:\sqrt{\mathrm{5}} \\ $$

Commented by peter frank last updated on 27/Apr/20

$$\sqrt{\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}\:}\:={a}+{b}\: \\ $$$$\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{5}}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab} \\ $$$$….. \\ $$

Commented by mathmax by abdo last updated on 28/Apr/20

$$\sqrt{\mathrm{7}}={a}\:{and}\:\sqrt{\mathrm{5}}={b}\:\Rightarrow\mathrm{7}={a}^{\mathrm{2}} \:{and}\:\mathrm{5}={b}^{\mathrm{2}} \:\Rightarrow\:\mathrm{12}\:={a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:=\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}\:\Rightarrow \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} \:={a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} +\mathrm{2}{ab}\:=\mathrm{12}+\mathrm{2}\sqrt{\mathrm{35}}\:\Rightarrow{a}+{b}\:=\sqrt{\mathrm{12}+\mathrm{2}\sqrt{\mathrm{35}}} \\ $$$$\sqrt{{a}+{b}}=\left(\mathrm{12}+\mathrm{2}\sqrt{\mathrm{35}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:=^{\mathrm{4}} \sqrt{\mathrm{12}+\mathrm{2}\sqrt{\mathrm{35}}}\:{after}\:{we}\:{use}\:{the}\:{calculator}… \\ $$

Commented by MJS last updated on 28/Apr/20

$$\mathrm{yes}\:\mathrm{but}\:\sqrt{\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}}=\sqrt[{\mathrm{4}}]{\mathrm{12}+\mathrm{2}\sqrt{\mathrm{35}}}\:\mathrm{is}\:\mathrm{something}\:\mathrm{I} \\ $$$$\mathrm{would}\:\mathrm{call}\:\mathrm{anti}−\mathrm{simplification}.\:\mathrm{obviously} \\ $$$$\left(\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{12}+\mathrm{2}\sqrt{\mathrm{35}}.\:\mathrm{why}\:\mathrm{not}\:\mathrm{say}\:\sqrt{\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}}= \\ $$$$=\sqrt[{\mathrm{8}}]{\mathrm{284}+\mathrm{48}\sqrt{\mathrm{35}}}? \\ $$

Commented by Prithwish Sen 1 last updated on 28/Apr/20

$$\mathrm{or}\:\mathrm{just}\:\mathrm{another}\:\mathrm{way} \\ $$$$\sqrt{\mathrm{7}}\:=\:\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+\mathrm{7}\sqrt{\mathrm{1}+\mathrm{8}…}}}\:\:\mathrm{and} \\ $$$$\sqrt{\mathrm{5}}\:=\:\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}\sqrt{\mathrm{1}+\mathrm{6}….}}} \\ $$$$\therefore\:\sqrt{\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}}\:=\:\left[\sqrt{\mathrm{1}+\mathrm{6}\sqrt{\mathrm{1}+\mathrm{7}…}}\:+\sqrt{\mathrm{1}+\mathrm{4}\sqrt{\mathrm{1}+\mathrm{5}…}}\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

Commented by MJS last updated on 28/Apr/20

$$\mathrm{yes}! \\ $$

Commented by Prithwish Sen 1 last updated on 28/Apr/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by MJS last updated on 27/Apr/20

$$\approx\mathrm{4}.\mathrm{88181928856} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{exact}\:\mathrm{expression}\:\mathrm{than}\:\sqrt{\sqrt{\mathrm{7}}+\sqrt{\mathrm{5}}} \\ $$

Commented by I want to learn more last updated on 27/Apr/20

$$\mathrm{Ohh},\:\mathrm{i}\:\mathrm{have}\:\mathrm{been}\:\mathrm{trying}\:\mathrm{sir}. \\ $$

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