How do you simplify #sqrt(10a) * sqrt(30a)# ?

2 Answers
Nam D.
Jan 31, 2018

#10asqrt(3)#

Explanation:

#sqrt(10a)*sqrt(30a)#

Know that, #sqrt(a)*sqrt(b)=sqrt(ab)# for #a, b>0#

#:.=sqrt(10a*30a)#

#=sqrt(300a^2)#

#=asqrt(300)#

#=asqrt(100*3)#

#=10asqrt(3)#

George C.
Jan 31, 2018

#sqrt(10a) * sqrt(30a) = 10sqrt(3)a#

Explanation:

If #a >= 0# and #b# is any number (positive, zero, negative or complex) then:

#sqrt(ab) = sqrt(a)sqrt(b)#

Moreover, if #a# is any number, then #sqrt(a)# is by definition a value satisfying:

#(sqrt(a))^2 = a#

So we find:

#sqrt(10a) * sqrt(30a) = sqrt(10) * sqrt(a) * sqrt(30) * sqrt(a)#

#color(white)(sqrt(10a) * sqrt(30a)) = sqrt(10) * sqrt(a) * sqrt(10) * sqrt(3) * sqrt(a)#

#color(white)(sqrt(10a) * sqrt(30a)) = (sqrt(10))^2 * sqrt(3) * (sqrt(a))^2#

#color(white)(sqrt(10a) * sqrt(30a)) = 10sqrt(3)a#

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