Question

rotionalise ( 6/sqrt√2+sqrt√3)​

Answer 1

Solution:

We have to rationalise,

$\tt = \dfrac{6}{ \sqrt{2} + \sqrt{3} }$

By multiplying both numerator and denominator by the conjugate of denominator, we get,

$\tt = \dfrac{6 \times ( \sqrt{2} - \sqrt{3}) }{ (\sqrt{2} + \sqrt{3})( \sqrt{2} - \sqrt{3})}$

$\tt = \dfrac{6 \times ( \sqrt{2} - \sqrt{3}) }{ (\sqrt{2})^{2} - ( \sqrt{3})^{2}}$

$\tt = \dfrac{6 \times ( \sqrt{2} - \sqrt{3}) }{2-3}$

$\tt = \dfrac{6 \times ( \sqrt{2} - \sqrt{3}) }{ - 1}$

$\tt = - 6 \times ( \sqrt{2} - \sqrt{3})$

$\tt = - 6 \sqrt{2} + 6 \sqrt{3}$

$\tt = 6 \sqrt{3} - 6 \sqrt{2}$

Which is our required answer.

Answer:

• 6√3 - 6√2

Concept Used:

• Rationalization: It is a process to remove the surds from the denominator part of a fraction by multiplying both the numerator and denominator by the conjugate of denominator. For example, conjugate of √2 + √3 is √2 - √3.

•••♪