Question

rationalise the denominator of following:​

Answer 1

☆ ANSWER:-

Your answer is

$\tt\mapsto \frac{6 \sqrt{6} + 4 }{25} .$

In the given question we have to rationalize the denominator of $\frac{2\sqrt{2}}{3\sqrt{3}-\sqrt{2}}.$

Concept Used:-

$\\ \tt\leadsto \frac{p}{q} \times \frac{a}{a} \\ \\ \tt = \frac{p}{q} \times 1 \\ \\ =\tt \frac{p}{q} \\ \\ \tt \: so \: we \: can \: write \frac{p}{q} \\ \\ \tt as \frac{p}{q} \times \frac{a}{a} .$

EXPLANATION:-

So before Rationalizing let us first understand the meaning of rationalization:-

• Rationalization means converting an irrational number to a rational number.

• Therefore Rationalizing denominator means we have to convert the denominator to a rational number.

For This we will use an algebraic ID:-

$\\ \tt\leadsto (a + b)(a - b) = {a}^{2} - {b}^{2} .$

Now lets apply this ID to rationalize the denominator:-

$\\ \tt\mapsto \frac{2 \sqrt{2} }{3 \sqrt{3} - \sqrt{2} } \\ \\ \tt = \frac{2 \sqrt{2} }{3 \sqrt{3} - \sqrt{2} } \times \frac{3 \sqrt{3} + \sqrt{2} }{ 3\sqrt{3} + \sqrt{2} } \\ \\ \tt = \frac{(2 \sqrt{2})(3 \sqrt{3} + \sqrt{2} ) }{(3 \sqrt{3} - \sqrt{2})(3 \sqrt{3} + \sqrt{2} ) } \\ \\ \tt = \frac{6 \sqrt{6} + 4 }{(3 \sqrt{3}) {}^{2} - ( \sqrt{2} ) {}^{2} } \\ \\ \tt = \frac{6 \sqrt{6} + 4 }{27 - 2} \\ \\ \tt = \frac{6 \sqrt{6} + 4 }{25}.$

Se we can see that denominator is a rational number hence the denominator is rationalized.

Here I am solving numerator separately so that you can understand betterly:-

$\\ \\ \sf \mapsto2 \sqrt{2}(3 \sqrt{3} + \sqrt{2}) \\ \\ \sf= (2\sqrt{2} \times 3 \sqrt{3} ) + ( 2\sqrt{2} \times \sqrt{2} ) \\ \\ \sf = (2 \times 3 \times \sqrt{3} \times \sqrt{2} ) + (2 \times \sqrt{2} \times \sqrt{2} ) \\ \\ \sf = (6 \times \sqrt{6} )+ (2 \times 2) \\ \\ \sf = 6 \sqrt{6} + 4.$