What is the possible answer for #sqrt2x(sqrt8x-sqrt32)#? How to simplify the answer too?

Question

#sqrt2x(sqrt8x-sqrt32)#

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Answer 1 · 2017-09-26T11:16:56

#sqrt(2) sqrt(x) (2sqrt(2)sqrt(x) - 4sqrt(2))#

#color(red)(root(n)(ab) = root(n)(a) * root(n)(b)) #

#sqrt(2x)# must have been the result of:
#sqrt(2) * sqrt(x)#

Now that's out of the way, using the same logic:
How did they get #sqrt(8x)# ?

Pull it apart and you get:
#sqrt(8) = 2sqrt(2)# and #sqrt(x)#

Same thing here: #sqrt(32)# = #4sqrt(2)#

After picking apart everything we get:

#color(red)(sqrt(2x)(sqrt(8x) - sqrt(32))) = ...#
#sqrt(2) sqrt(x) (2sqrt(2)sqrt(x) - 4sqrt(2))#

Simplifying:
#color(red)(a(b+c) = ab+ac#

#(sqrt(2)sqrt(x) * 2sqrt(2)sqrt(x)) - (sqrt(2)sqrt(x) * 4sqrt(2))#

#sqrt(2)sqrt(x) * 2sqrt(2)sqrt(x) = 4x#

#sqrt(2)sqrt(x) * 4sqrt(2) = 8sqrt(x)#

#4x - 8sqrt(x)#

Answer 2 · 2017-09-26T11:31:43

Given
#sqrt(2) x (sqrt(8)x - sqrt(32))#

Let us take #sqrt2# inside the parentheses and multiply both terms. It becomes

#x (sqrt2xxsqrt8x - sqrt2xxsqrt(32))#
#=>x (sqrt(8xx2)x - sqrt(32xx2))#
#=>x (sqrt(16)x - sqrt(64))#
#=>x (4x - 8)#

Taking common factor #4# outside the parentheses we get simplified form as

#4x (x - 2)#