Question

$\sqrt{125 } - 4 \sqrt{6} + \sqrt{294} - 2 \sqrt{ \frac{1}{6} }$
simplify​

Answer 1

Solution!!

$\sf \sqrt{125}-4\sqrt{6}+\sqrt{294}-2\sqrt{\dfrac{1}{6}}$

Simplify the radical expressions

$\sf = \sqrt{5^{3}}-4\sqrt{6}+\sqrt{7^{2}\times 6}-2\sqrt{\dfrac{1}{6}}$

$\sf = \sqrt{5^{2+1}}-4\sqrt{6}+\sqrt{7^{2}}\sqrt{6}-2\sqrt{\dfrac{1}{6}}$

$\sf = \sqrt{5^{2}\times 5^{1}}-4\sqrt{6}+7\sqrt{6}-2\sqrt{\dfrac{1}{6}}$

$\sf = \sqrt{5^{2}}\sqrt{5}-4\sqrt{6}+7\sqrt{6}-2\sqrt{\dfrac{1}{6}}$

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-2\sqrt{\dfrac{1}{6}}$

To take the root of a fraction, take the root of numerator and denominator separately.

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-2\times \dfrac{1}{\sqrt{6}}$

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-\dfrac{2}{\sqrt{6}}$

Rationalise the denominator

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-\dfrac{2}{\sqrt{6}}\times \dfrac{\sqrt{6}}{\sqrt{6}}$

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-\dfrac{2\sqrt{6}}{\sqrt{6}\times \sqrt{6}}$

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-\dfrac{2\sqrt{6}}{6}$

$\sf = 5\sqrt{5}-4\sqrt{6}+7\sqrt{6}-\dfrac{\sqrt{6}}{3}$

Find the sum and the difference.

$\sf = 5\sqrt{5}+\dfrac{-12\sqrt{6}+21\sqrt{6}-\sqrt{6}}{3}$

$\sf = 5\sqrt{5}+\dfrac{-13\sqrt{6}+21\sqrt{6}}{3}$

$\boxed{\sf = 5\sqrt{5}+\dfrac{8\sqrt{6}}{3}}$