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

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

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

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

\sqrt{\sqrt{7} + \sqrt{5}} = a + b

\sqrt{7} + \sqrt{5} = a^{2} + b^{2} + 2 a b

\dots . .

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

\sqrt{7} = a a n d \sqrt{5} = b \Rightarrow 7 = a^{2} a n d 5 = b^{2} \Rightarrow 12 = a^{2} + b^{2} = {(a + b)}^{2} - 2 a b \Rightarrow

{(a + b)}^{2} = a^{2} + b^{2} + 2 a b = 12 + 2 \sqrt{35} \Rightarrow a + b = \sqrt{12 + 2 \sqrt{35}}

\sqrt{a + b} = {(12 + 2 \sqrt{35})}^{\frac{1}{4}} =^{4} \sqrt{12 + 2 \sqrt{35}} a f t e r w e u s e t h e c a l c u l a t o r \dots

Commented by MJS last updated on 28/Apr/20

yes but \sqrt{\sqrt{7} + \sqrt{5}} = \sqrt[4]{12 + 2 \sqrt{35}} is something I

would call anti - simplification . obviously

{(\sqrt{7} + \sqrt{5})}^{2} = 12 + 2 \sqrt{35} . why not say \sqrt{\sqrt{7} + \sqrt{5}} =

= \sqrt[8]{284 + 48 \sqrt{35}} ?

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

or just another way

\sqrt{7} = \sqrt{1 + 6 \sqrt{1 + 7 \sqrt{1 + 8 \dots}}} and

\sqrt{5} = \sqrt{1 + 4 \sqrt{1 + 5 \sqrt{1 + 6 \dots .}}}

∴ \sqrt{\sqrt{7} + \sqrt{5}} = {[\sqrt{1 + 6 \sqrt{1 + 7 \dots}} + \sqrt{1 + 4 \sqrt{1 + 5 \dots}}]}^{\frac{1}{2}}

Commented by MJS last updated on 28/Apr/20

yes!

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

thank you sir .

Answered by MJS last updated on 27/Apr/20

\approx 4.88181928856

no other exact expression than \sqrt{\sqrt{7} + \sqrt{5}}

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

Ohh, i have been trying sir .