Question Number 59861 by ANTARES VY last updated on 15/May/19
$$\sqrt{\boldsymbol{{a}}+\boldsymbol{{b}}\sqrt{\boldsymbol{{c}}}}=\sqrt{\frac{\boldsymbol{{a}}+\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}+\sqrt{\frac{\boldsymbol{{a}}−\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}. \\ $$$$\boldsymbol{{prove}} \\ $$
Commented by Kunal12588 last updated on 15/May/19
Commented by Kunal12588 last updated on 15/May/19
$${just}\:{put}\:“\:{b}^{\mathrm{2}} {c}\:''\:{in}\:{place}\:{of}\:“\:{b}\:'' \\ $$
Answered by tanmay last updated on 15/May/19
![((a+(√(a^2 −b^2 c)))/2) =((2a+2(√((a+b(√c) )(a−b(√c) ))))/4) =((a+b(√c) +a−b(√c) +2(√((a+b(√(c)(a−b(√(c))) )))))/4) =((((√(a+b(√c) )) )^2 +((√(a−b(√c) )) )^2 +2(√((a+b(√c) )(a−b(√(c))) )))/4) =(([(√(a+b(√c))) +(√(a−b(√c) )) ]^2 )/2^2 ) so (√((a+(√(a^2 −b^2 c)))/2)) +(√((a−(√(a^2 −b^2 c)))/2)) =(((√(a+b(√c) )) +(√(a−b(√c))) )/2)+(((√(a+b(√c))) −(√(a−b(√c))) )/2) =(√(a+b(√c)))](https://www.tinkutara.com/question/Q59862.png)
$$\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2}{a}+\mathrm{2}\sqrt{\left({a}+{b}\sqrt{{c}}\:\right)\left({a}−{b}\sqrt{{c}}\:\right)}}{\mathrm{4}} \\ $$$$=\frac{{a}+{b}\sqrt{{c}}\:+{a}−{b}\sqrt{{c}}\:+\mathrm{2}\sqrt{\left({a}+{b}\sqrt{\left.{c}\right)\left({a}−{b}\sqrt{\left.{c}\right)}\:\right.}\right.}}{\mathrm{4}} \\ $$$$=\frac{\left(\sqrt{{a}+{b}\sqrt{{c}}\:}\:\right)^{\mathrm{2}} +\left(\sqrt{{a}−{b}\sqrt{{c}}\:}\:\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\left({a}+{b}\sqrt{{c}}\:\right)\left({a}−{b}\sqrt{\left.{c}\right)}\:\right.}}{\mathrm{4}} \\ $$$$=\frac{\left[\sqrt{{a}+{b}\sqrt{{c}}}\:+\sqrt{{a}−{b}\sqrt{{c}}\:\:}\:\right]^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} } \\ $$$${so} \\ $$$$\sqrt{\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}}}{\mathrm{2}}}\:+\sqrt{\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}}}{\mathrm{2}}}\: \\ $$$$=\frac{\sqrt{{a}+{b}\sqrt{{c}}\:}\:+\sqrt{{a}−{b}\sqrt{{c}}}\:}{\mathrm{2}}+\frac{\sqrt{{a}+{b}\sqrt{{c}}}\:−\sqrt{{a}−{b}\sqrt{{c}}}\:}{\mathrm{2}} \\ $$$$=\sqrt{{a}+{b}\sqrt{{c}}}\: \\ $$$$ \\ $$
Answered by Kunal12588 last updated on 15/May/19
$${let}\:\sqrt{{a}+{b}\sqrt{{c}}}=\sqrt{{m}}+\sqrt{{n}}\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow{a}+\sqrt{{b}^{\mathrm{2}} {c}}={m}+{n}+\mathrm{2}\sqrt{{mn}} \\ $$$$\Rightarrow{m}+{n}={a}\:,\:\mathrm{4}{mn}={b}^{\mathrm{2}} {c} \\ $$$$\Rightarrow{m}−{n}=\sqrt{\left({m}+{n}\right)^{\mathrm{2}} −\mathrm{4}{mn}}=\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}} \\ $$$$\Rightarrow{m}=\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}}}{\mathrm{2}}\:,\:{n}=\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}}}{\mathrm{2}} \\ $$$${from}\:\left(\mathrm{1}\right) \\ $$$$\sqrt{{a}+{b}\sqrt{{c}}}=\sqrt{{m}}+\sqrt{{n}} \\ $$$$\Rightarrow\sqrt{\boldsymbol{{a}}+\boldsymbol{{b}}\sqrt{\boldsymbol{{c}}}}=\sqrt{\frac{\boldsymbol{{a}}+\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}+\sqrt{\frac{\boldsymbol{{a}}−\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}} \\ $$$$\boldsymbol{{proved}} \\ $$