Question

Simplify 2 root 3 by 3 minus root 3 by 6

Answer 1

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Answer 2

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=6\sqrt3+6$

Step-by-step explanation:

Given : Expression $\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}$

To find : Simplify the expression ?

Solution :

Writing the expression as,

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=\frac{12\sqrt3}{3-\sqrt3}$

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=\frac{12\sqrt3}{3-\sqrt3}\times\frac{3+\sqrt3}{3+\sqrt3}$

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=\frac{(12\sqrt3)(3+\sqrt3)}{3^2-(\sqrt3)^2}$

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=\frac{(12\sqrt3)(3+\sqrt3)}{9-3}$

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=\frac{(12\sqrt3)(3+\sqrt3)}{6}$

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=2\sqrt3(3+\sqrt3)$

$\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=6\sqrt3+6$

Therefore, $\frac{2\sqrt3}{\frac{3-\sqrt3}{6}}=6\sqrt3+6$

#Learn more

Simplify (√5+√3) (√5-√3)

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