Question

find the value of the above question​

Answer 1

Let,

$\rm\sqrt{25+10\sqrt{6}} + \sqrt{25-10\sqrt{6}} = x$

Squaring both the sides,

$\rm\dashrightarrow\bigg (\sqrt{25+10\sqrt{6}} + \sqrt{25-10\sqrt{6}} \bigg )^2= x^2$

We know that,

(a + b)² = a² + b² + 2ab

Hence,

$\rm\dashrightarrow\bigg (\sqrt{25+10\sqrt{6}} \bigg )^2 + \bigg ( \sqrt{25-10\sqrt{6}} \bigg )^2+2\sqrt{25+10\sqrt{6}}\sqrt{25-10\sqrt{6}}= x^2$

$\rm\dashrightarrow 25+10\sqrt{6}+ 25-10\sqrt{6}+2\sqrt{25+10\sqrt{6}}\sqrt{25-10\sqrt{6}}= x^2$

We know that,

√a√b = √ab

Hence,

$\rm\dashrightarrow 25+10\sqrt{6}+25-10\sqrt{6}+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}= x^2$

• -10√6 and 10√6 get cancelled.

Take 5 as common factor in 25+10√6 as well as 25-10√6

$\rm\dashrightarrow 25+25+2\sqrt{5(5+2\sqrt{6})5(5-2\sqrt{6})}= x^2$

$\rm\dashrightarrow 50+2\sqrt{25[(5+2\sqrt{6})(5-2\sqrt{6})]}= x^2$

• a² - b² = (a + b)(a - b)

$\rm\dashrightarrow 50+2\sqrt{25[(5)^2-(2\sqrt{6})^2]}= x^2$

$\rm\dashrightarrow 50+2\sqrt{25(25-24)}= x^2$

$\rm\dashrightarrow 50+2\sqrt{25(1)}= x^2$

$\rm\dashrightarrow 50+2\sqrt{25}= x^2$

• √25 = 5

$\rm\dashrightarrow 50+2(5)= x^2$

$\rm\dashrightarrow 50+10= x^2$

$\rm\dashrightarrow 60= x^2$

$\rm\dashrightarrow \sqrt{60}= x$

$\rm\dashrightarrow \sqrt{2 \times 2 \times 15}= x$

$\rm\dashrightarrow 2\sqrt{15}= x$

Hence,

$\bf\dashrightarrow x= 2\sqrt{15}$

Hence, simplified.