Question

answer this.....,. ​

Answer 1

Question:-

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

then find the value of x.

Solution:-

$\sf \implies \dfrac{(x - 2\sqrt{6})(5\sqrt{3} + 5\sqrt{2})}{5\sqrt{3} - 5\sqrt{2}} = 1$

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

$\sf \implies x - 2\sqrt{6} = \dfrac{5\sqrt{3} - 5\sqrt{2}}{5\sqrt{3} + 5\sqrt{2}}$

By rationalizing the denominator:-

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

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

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

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

$\sf \implies x - 2\sqrt{6} = \dfrac{125 - 50 \sqrt{6} }{25}$

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

$\sf \implies x - 2\sqrt{6} = 5 - 2 \sqrt{6}$

Comparing LHS and RHS:-

$\sf \implies x = 5$

$\sf \large\dashrightarrow \underline{ \boxed{ \therefore \textsf{ \textbf{The \: value \: of \: x = 5}} }}$

$\sf \underline{\:\: \underline{\: Option \: (4) \: is \: correct \:}\:\:}$