Question

Rationalise the denominator of:(3√2 - 2√3)/(√8 + √27)​

Answer 1

Step-by-step explanation:

$\large{ \sf \underline{Solution - }}$

$\sf To \: rationalise \: = \: \dfrac{3 \sqrt{2} - 2 \sqrt{3}}{\sqrt{8} + \sqrt{27} }$

$\sf \dashrightarrow \dfrac{3 \sqrt{2} - 2 \sqrt{3}}{2 \sqrt{2} + 3 \sqrt{3} }$

$\sf \dashrightarrow \dfrac{3 \sqrt{2} - 2 \sqrt{3}}{2 \sqrt{2} + 3 \sqrt{3} } \times \dfrac{2 \sqrt{2} - 3 \sqrt{3}}{2 \sqrt{2} - 3 \sqrt{3}}$

$\sf Combine \: the \: fractions,$

$\sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{(2 \sqrt{2} + 3 \sqrt{3} )(2 \sqrt{2} - 3 \sqrt{3})}$

$\sf \underline{Solving \: denominator} ,$

$\sf We \: know \: that,$

$\sf \dashrightarrow \boxed{ \tt (a + b)(a - b) = a^{2} - b ^{2}}$

$\bf So ,$

$\sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{(2 \sqrt{2})^{2} - (3 \sqrt{3} )^{2} }$

$\sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{8 - 27 }$

$\sf \dashrightarrow \dfrac{(3 \sqrt{2} - 2 \sqrt{3})(2 \sqrt{2} - 3 \sqrt{3})}{ - 19 }$

$\sf \underline{Solving \: numerator},$

$\sf \dashrightarrow \dfrac{3 \sqrt{2} (2 \sqrt{2} - 3 \sqrt{3})- 2 \sqrt{3}(2 \sqrt{2} - 3 \sqrt{3})}{ - 19 }$

$\sf \dashrightarrow \dfrac{12 - 9 \sqrt{6} - 4 \sqrt{6} + 18}{ - 19 }$

$\sf \dashrightarrow \dfrac{30 - 13 \sqrt{6} }{ - 19 }$

$\sf \dashrightarrow - \dfrac{30 - 13 \sqrt{6} }{ 19 }$

$\bf Hence,$

$\sf On \: rationalising \: we \: got,$

$\bf \dashrightarrow - \dfrac{30 - 13 \sqrt{6} }{ 19 }$