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

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Answer 1 · 2018-04-13T12:14:42

$21sqrt2 + 6sqrt6, or 3(7sqrt2 + 2sqrt6)$

the square root of $6$ can be written as $sqrt6$ .

$7$ multiplied by the square root of $3$ can be written as $7sqrt3$ .

$6$ added to $7$ multiplied by the square root of $3$ can be written as $7sqrt3 + 6$

therefore the square root of $6 *$ ( $7$ multiplied by the square root of $3$ ) $+ 6$ ) is written as $sqrt6(7sqrt3+6)$ .

to solve $sqrt6(7sqrt3+6)$ , multiply the two terms in the bracket separately with the term outside the bracket.

$sqrt6 * 7sqrt3 = 7 * (sqrt6 * sqrt3) = 7 sqrt18$

$sqrt18 = sqrt9 * sqrt2 = 3 * sqrt2$

$7 * sqrt18 = 7 * 3 * sqrt2 = 21 * sqrt2$

$sqrt6 * 7sqrt3 = 21sqrt2$

$sqrt6 * 6 = 6sqrt6$

$sqrt6(7sqrt3+6) = (sqrt6 * 7sqrt3) + (sqrt6 * 6)$

$= 21sqrt2 + 6sqrt6$

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

$21sqrt2 = 3 * 7sqrt2$

$6sqrt6 = 3 * 2sqrt6$

$21sqrt2 + 6sqrt6 = 3(7sqrt2 + 2sqrt6)$