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Question Number 199837 by Calculusboy last updated on 10/Nov/23

Answered by deleteduser1 last updated on 10/Nov/23

$$\sqrt{a+\sqrt{b}}=p;\sqrt{a-\sqrt{b}}=q\Rightarrow p^2+q^2=2a;p^2{q}^2=a^2-b$$$$\Rightarrow pq=\sqrt{a^2-b}\Rightarrow {(p+q)}^2=2a+2\sqrt{a^2-b}$$$$\Rightarrow p+q=\sqrt{2}(a+\sqrt{a^2-b})$$$$\Rightarrow p,qarerootsof$$$$x^2-\sqrt{2}\left(\sqrt{a+\sqrt{a^2-b}}\right)x+\sqrt{a^2-b}$$$$\Rightarrow p,q=\frac{\sqrt{2}\left(\sqrt{a+\sqrt{a^2-b}}\underset{-}{+}\sqrt{a-\sqrt{a^2-b}}\right)}{2}$$$$\Rightarrow p=a+\sqrt{b}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$

Commented by Calculusboy last updated on 11/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

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