# What is the square root of 6 (7 the square root of 3 + 6)?

Apr 13, 2018

$21 \sqrt{2} + 6 \sqrt{6} , \mathmr{and} 3 \left(7 \sqrt{2} + 2 \sqrt{6}\right)$

#### Explanation:

the square root of $6$ can be written as $\sqrt{6}$.

$7$ multiplied by the square root of $3$ can be written as $7 \sqrt{3}$.

$6$ added to $7$ multiplied by the square root of $3$ can be written as $7 \sqrt{3} + 6$

therefore the square root of $6 \cdot$ ($7$ multiplied by the square root of $3$)$+ 6$) is written as $\sqrt{6} \left(7 \sqrt{3} + 6\right)$.

to solve $\sqrt{6} \left(7 \sqrt{3} + 6\right)$, multiply the two terms in the bracket separately with the term outside the bracket.

$\sqrt{6} \cdot 7 \sqrt{3} = 7 \cdot \left(\sqrt{6} \cdot \sqrt{3}\right) = 7 \sqrt{18}$

$\sqrt{18} = \sqrt{9} \cdot \sqrt{2} = 3 \cdot \sqrt{2}$

$7 \cdot \sqrt{18} = 7 \cdot 3 \cdot \sqrt{2} = 21 \cdot \sqrt{2}$

$\sqrt{6} \cdot 7 \sqrt{3} = 21 \sqrt{2}$

$\sqrt{6} \cdot 6 = 6 \sqrt{6}$

$\sqrt{6} \left(7 \sqrt{3} + 6\right) = \left(\sqrt{6} \cdot 7 \sqrt{3}\right) + \left(\sqrt{6} \cdot 6\right)$

$= 21 \sqrt{2} + 6 \sqrt{6}$

the roots cannot be simplified further, but you may wish to factorise:

$21 \sqrt{2} = 3 \cdot 7 \sqrt{2}$

$6 \sqrt{6} = 3 \cdot 2 \sqrt{6}$

$21 \sqrt{2} + 6 \sqrt{6} = 3 \left(7 \sqrt{2} + 2 \sqrt{6}\right)$