Question

$If \: \frac{(x - 2\sqrt{6})(5\sqrt{3} + 5\sqrt{2})}{5\sqrt{3} - 5\sqrt{2}} = 1$

then find the value of x.​

Answer 1

x=5

Step by step solution is given in the picture

Answer 2

$\qquad$ Solution:-

$\pink{\qquad\leadsto\quad\sf \dfrac{(x - 2\sqrt{6})(5\sqrt{3} + 5\sqrt{2})}{5\sqrt{3} - 5\sqrt{2}} = 1}$

$\qquad\leadsto\quad\sf (x - 2\sqrt{6})(5\sqrt{3} + 5\sqrt{2}) = 5\sqrt{3} - 5\sqrt{2}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{5\sqrt{3} - 5\sqrt{2}}{5\sqrt{3} + 5\sqrt{2}}$

$\qquad$ ㅤ$\small {\mathfrak {\underline {By \: rationalizing\: the \: denominator:- }}}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{5\sqrt{3} - 5\sqrt{2}}{5\sqrt{3} + 5\sqrt{2}} \times \dfrac{5\sqrt{3} - 5\sqrt{2}}{5\sqrt{3} - 5\sqrt{2}}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{(5\sqrt{3} - 5\sqrt{2})^{2} }{(5\sqrt{3})^{2} - (5\sqrt{2})^{2} }$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{(5\sqrt{3})^{2} + (5\sqrt{2})^{2} - 2(5 \sqrt{3})(5 \sqrt{2})}{75 - 50}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{75 + 50 - 50 \sqrt{6} }{25}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{125 - 50 \sqrt{6} }{25}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = \dfrac{25(5 - 2 \sqrt{6}) }{25}$

$\qquad\leadsto\quad\sf x - 2\sqrt{6} = 5 - 2 \sqrt{6}$

$\qquad\leadsto\quad\sf x = 5 - 2 \sqrt{6} + 2\sqrt{6}$

$\qquad\leadsto\quad\sf x = 5 - \cancel{ 2 \sqrt{6} }+ \cancel{2\sqrt{6}}$

$\qquad \leadsto \underline {\boxed{\pmb{\frak{\pink{ x\:=\:5 \:}}}}}\:\:\bigstar$

$\therefore\:\underline{\textsf{Value of x is \textbf{5}}}.$