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Question Number 59777 by Kunal12588 last updated on 14/May/19

Commented by Kunal12588 last updated on 14/May/19

$$Imadeaformulayoucanuseitlikethis$$$$\sqrt{2+\sqrt{3}}$$$$1.findk$$$$k=a-\sqrt{a^2-b}\left[\sqrt{2+\sqrt{3}}\right]$$$$\Rightarrow k=2-\sqrt{4-3}=1$$$$2.\sqrt{2+\sqrt{3}}=\sqrt{\frac{1}{2}}(1+\frac{\sqrt{3}}{1})=\frac{1+\sqrt{3}}{\sqrt{2}}$$$$CanAnyonepleasetrymyformula$$

Commented by Kunal12588 last updated on 14/May/19

$$\sqrt{2-\sqrt{3}}$$$$1.findk$$$$k=a+\sqrt{a^2-b}\left[\sqrt{2-\sqrt{3}}\right]$$$$k=2+1=3$$$$2.\sqrt{2-\sqrt{3}}=\sqrt{\frac{3}{2}}(1-\frac{\sqrt{3}}{3})=\sqrt{\frac{3}{2}}(1-\frac{1}{\sqrt{3}})=\frac{\sqrt{3}-1}{\sqrt{2}}$$

Commented by MJS last updated on 14/May/19

$$\mathrm{seems}\mathrm{to}\mathrm{work}$$$$\mathrm{but}\mathrm{only}\mathrm{if}\sqrt{a^2-b}\in \mathbb{N}\mathrm{otherwise}\mathrm{it}{\mathrm{doesn}}^{\prime }\mathrm{t}$$$$\mathrm{help}$$

Commented by Kunal12588 last updated on 15/May/19

$$thankyousir$$

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