Question

Rationdisc: 3÷2√5-3√2​

Answer 1

Question:-

Rationalise the denominator

$\sf{\dfrac{3}{2\sqrt{5} - 3\sqrt{2}}}$

Solution:-

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

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

= $\sf{\dfrac{3(2\sqrt{5} + 3\sqrt{2})}{20-19}}$

= $\sf{\dfrac{6\sqrt{5} + 9\sqrt{2}}{1}}$

= $\sf{6\sqrt{5} + 9\sqrt{2}}$

What is rationalising the denominator?

Answer- Rationalising a denominator is the process of converting the irrational denominator into it's rational form.

For example,

$\sf{Let \:us\:consider\:a\:fraction\dfrac{2}{\sqrt{3} - \sqrt{2}}}$

In this we have an irrational denominator.

We need to convert it into its rational form.

Therefore,

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

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

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

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

= $\sf{\dfrac{2\sqrt{3}+2\sqrt{2}}{1}}$

= $\sf{2\sqrt{3}+2\sqrt{2}}$

So we converted the irrational denominator $\sf{\dfrac{2}{\sqrt{3}-\sqrt{2}}}$ into it's rational form i.e., 1.