Question

find the value of

$\sqrt{15 + 10 \sqrt{2} } + \sqrt{15 - 10 \sqrt{2 } }$
​

Answer 1

$\text{Consider,}$

$\bf\displaystyle\,15+10\sqrt{2}$

$=\displaystyle\,10+5+2(5)\sqrt{2}$

$=\displaystyle\,(\sqrt{10})^2+(\sqrt{5})^2+2(\sqrt{5})(\sqrt{5})\sqrt{2}$

$=\displaystyle\,(\sqrt{10})^2+(\sqrt{5})^2+2(\sqrt{10})(\sqrt{5})$

$\text{Using, }$

$\boxed{\bf\,a^2+b^2+2ab=(a+b)^2}$

$=\displaystyle\,(\sqrt{10}+\sqrt{5})^2$

$\implies\displaystyle\,15+10\sqrt{2}=(\sqrt{10}+\sqrt{5})^2$

$\implies\bf\displaystyle\sqrt{15+10\sqrt{2}}=\sqrt{10}+\sqrt{5}$ .....(1)

$\bf\displaystyle\,15-10\sqrt{2}$

$=\displaystyle\,10+5-2(5)\sqrt{2}$

$=\displaystyle\,(\sqrt{10})^2+(\sqrt{5})^2-2(\sqrt{5})(\sqrt{5})\sqrt{2}$

$=\displaystyle\,(\sqrt{10})^2+(\sqrt{5})^2-2(\sqrt{10})(\sqrt{5})$

$\text{Using, }$

$\boxed{\bf\,a^2+b^2-2ab=(a-b)^2}$

$=\displaystyle\,(\sqrt{10}-\sqrt{5})^2$

$\implies\displaystyle\,15+10\sqrt{2}=(\sqrt{10}-\sqrt{5})^2$

$\implies\bf\displaystyle\sqrt{15-10\sqrt{2}}=\sqrt{10}-\sqrt{5}$ ......(2)

$\text{Adding (1) and (2), we get}$

$\displaystyle\sqrt{15+10\sqrt{2}}+\sqrt{15-10\sqrt{2}}$

$=\sqrt{10}+\sqrt{5}+\sqrt{10}-\sqrt{5}$

$=2\sqrt{10}$

$\therefore\textbf{The value of \bf\sqrt{15+10\sqrt{2}}+\sqrt{15-10\sqrt{2}} is \bf\,2\sqrt{10} }$

Answer 2

Answer:

After dividing both side root 2 so answer will be =4.975