Question Number 199837 by Calculusboy last updated on 10/Nov/23
Answered by AST last updated on 10/Nov/23
$$\sqrt{{a}+\sqrt{{b}}}={p};\sqrt{{a}−\sqrt{{b}}}={q}\Rightarrow{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{2}{a};{p}^{\mathrm{2}} {q}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b} \\ $$$$\Rightarrow{pq}=\sqrt{{a}^{\mathrm{2}} −{b}}\Rightarrow\left({p}+{q}\right)^{\mathrm{2}} =\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} −{b}} \\ $$$$\Rightarrow{p}+{q}=\sqrt{\mathrm{2}}\left({a}+\sqrt{{a}^{\mathrm{2}} −{b}}\right) \\ $$$$\Rightarrow{p},{q}\:{are}\:{roots}\:{of}\: \\ $$$${x}^{\mathrm{2}} −\sqrt{\mathrm{2}}\left(\sqrt{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}\right){x}+\sqrt{{a}^{\mathrm{2}} −{b}} \\ $$$$\Rightarrow{p},{q}=\frac{\sqrt{\mathrm{2}}\left(\sqrt{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}\underset{−} {+}\sqrt{{a}−\sqrt{{a}^{\mathrm{2}} −{b}}}\right)}{\mathrm{2}} \\ $$$$\Rightarrow{p}={a}+\sqrt{{b}}=\sqrt{\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}}+\sqrt{\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}} \\ $$
Commented by Calculusboy last updated on 11/Nov/23
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$