How do you simplify #sqrt(3x^2 - 12x + 12)#?

1 Answer
George C.
Jul 13, 2015

#sqrt(3x^2-12x+12) = sqrt(3)*abs(x-2)#

Explanation:

When #a, b >= 0# we have #sqrt(ab) = sqrt(a)sqrt(b)#

#sqrt(3x^2-12x+12)#

#=sqrt(3*(x^2-4x+4))#

#=sqrt(3*(x-2)^2)#

Now #3 >= 0# and #(x-2)^2 >= 0# for all #x in RR#

So:

#sqrt(3*(x-2)^2)#

#= sqrt(3)*sqrt((x-2)^2)#

#= sqrt(3)*abs(x-2)#

Note that #(x-2)^2# has two square roots, namely #(x-2)# and #-(x-2)#.

#sqrt((x-2)^2)# denotes the positive square root, so we have to pick the positive value from #(x-2)# and #-(x-2)#, which we can do by using #abs(x-2)#

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