Question

rotationalise the denominator 2 square root 3+4 square root 5divide by 3 square root 2-square root 5

Answer 1

Correct Question:

Rationalise the denominator -

$\bf \dfrac{2\sqrt{3} + 4\sqrt{5} }{3\sqrt{2} - \sqrt{5}}$

Given:

A number -

$\bf \dashrightarrow \: \dfrac{2\sqrt{3} + 4\sqrt{5} }{3\sqrt{2} - \sqrt{5}}$

What To Find:

We have to -

• Rationalise the denominator.

How To Find:

To find we have to -

• First, we have to take the rationalising factor of the denominator.
• Next, we have to multiply that rationalising factor with both denominator and numerator.
• Then, simplify them using identities if possible.
• Final, the denominator is rationalised.

Solution:

Here, the rationalising factor is,

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

Multiply with both numerator and denominator,

$\sf \implies \dfrac{2\sqrt{3} + 4\sqrt{5} }{3\sqrt{2} - \sqrt{5}} \times \dfrac{3\sqrt{2} + \sqrt{5}}{3\sqrt{2} + \sqrt{5}}$

Take them as common,

$\sf \implies \dfrac{2\sqrt{3} + 4\sqrt{5} \times 3\sqrt{2} + \sqrt{5}}{3\sqrt{2} - \sqrt{5} \times 3\sqrt{2} + \sqrt{5}}$

Solve the numerator,

$\sf \implies \dfrac{2\sqrt{3}(3\sqrt{2} + \sqrt{5}) + 4\sqrt{5}(3\sqrt{2} + \sqrt{5})}{3\sqrt{2} - \sqrt{5} \times 3\sqrt{2} + \sqrt{5}}$

Further, solve the numerator,

$\sf \implies \dfrac{6\sqrt{6} + 2\sqrt{15} + 12\sqrt{10} + 20}{3\sqrt{2} - \sqrt{5} \times 3\sqrt{2} + \sqrt{5}}$

Using the identity (a -  b) (a + b) = a² - b² in the denominator,

$\sf \implies \dfrac{6\sqrt{6} + 2\sqrt{15} + 12\sqrt{10} + 20}{(3\sqrt{2})^2 - (\sqrt{5})^2}$

Remove the brackets,

$\sf \implies \dfrac{6\sqrt{6} + 2\sqrt{15} + 12\sqrt{10} + 20}{3\sqrt{2} \times 3\sqrt{2} - \sqrt{5} \times \sqrt{5}}$

Multiply them first,

$\sf \implies \dfrac{6\sqrt{6} + 2\sqrt{15} + 12\sqrt{10} + 20}{18 - 5}$

Subtract them next,

$\sf \implies \dfrac{6\sqrt{6} + 2\sqrt{15} + 12\sqrt{10} + 20}{13}$

∵ Here, 13 is a rational number,

∴ Hence, rationalised.

Final Answer:

∴ Thus, the answer is $\bf \dfrac{6\sqrt{6} + 2\sqrt{15} + 12\sqrt{10} + 20}{13}$ after rationalising the denominator.