Question

rationalise 6/3+2root2 + 5/3-2root2​

Answer 1

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

$\sf \longmapsto \dfrac{6}{3 + 2 \sqrt{2} } + \dfrac{5}{3 - 2 \sqrt{2} }$

Solving separately a part.

$\sf \longmapsto \dfrac{6}{3 + 2 \sqrt{2} } \times \dfrac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} }$

Identity used here :-

(a+b)(a-b)=a²-b²

$\sf \longmapsto \dfrac{6(3 - 2 \sqrt{2} )}{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} }$

$\sf \longmapsto \dfrac{6(3 - 2 \sqrt{2} )}{9 - 6} = \dfrac{6(3 - 2 \sqrt{2}) }{3}$

$\sf \bold{ = 2(3 - 2 \sqrt{2} )}$

$\sf \longmapsto \dfrac{5}{3 - 2 \sqrt{2} } \times \dfrac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }$

$\sf \longmapsto \dfrac{5(3 + 2 \sqrt{2} )}{ {3)}^{2} - {(2 \sqrt{2} )}^{2} } = \dfrac{5(3 + 2 \sqrt{2} )}{9 - 8}$

$\sf \bold{= 5(3 + 2 \sqrt{2} )}$

Add both the values :

$\sf \longmapsto2(3 - 2 \sqrt{2} ) + 5(3 + 2 \sqrt{2} )$

$\sf \longmapsto6 - 4 \sqrt{2} + 15 + 10 \sqrt{2}$

$\sf \longmapsto \bold{ \red{21 + 6 \sqrt{2} }}$

Extra information=>

Rationalising means removing root values from its denominators and shifts towards numerator.

Method for rationalising:-

1. Multiply with the opposite sign values of denominator to both numerator and denominator.