How do you simplify #7sqrt(28)#?

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Answer 1 · 2018-03-05T19:19:05

You find perfect squares that are factors in the radical.

#√28#

#√4=2#

#7*2√7#

#14√7#

Answer 2 · 2018-03-05T19:53:24

#14sqrt(7)#

#28# is the same as #2xx14# witch in turn is the same as #2^2xx7#

Write as #7sqrt(2^2xx7)#

Taking the #2^2# 'outside' the square root changes it from #2^2" to "2#

#7sqrt(2^2xx7) = 7xx2xxsqrt(7)#

#14sqrt7#