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

1 Answer
Apr 13, 2018

Answer:

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

Explanation:

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)#