How do you multiply #xsqrt(10x)*7sqrt(15x)#?

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Answer 1 · 2018-03-03T20:14:31

The result is #35x^2sqrt6#.

Multiplication: #sqrta*sqrtb=sqrt(ab)#

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

For this problem, first, multiply the radicals (in blue) and their coefficients (in red) together:

#color(white)=color(red)xcolor(blue)sqrt(10x)*color(red)7color(blue)sqrt(15x)#

#=color(red)x*color(blue)sqrt(10x)*color(red)7*color(blue)sqrt(15x)#

#=color(red)x*color(red)7*color(blue)sqrt(10x)*color(blue)sqrt(15x)#

#=color(red)(7x)*color(blue)sqrt(10x)*color(blue)sqrt(15x)#

#=color(red)(7x)*color(blue)sqrt(10x*15x)#

#=color(red)(7x)*color(blue)sqrt(10*15*x*x)#

#=color(red)(7x)\*color(blue)sqrt(150*x*x)#

#=color(red)(7x)*color(blue)sqrt(150*x^2)#

Next, use the multiplication rule backwards:

#color(white)=color(red)(7x)*color(blue)sqrt(150*x^2)#

#=color(red)(7x)*color(blue)sqrt(150)*color(blue)(sqrt(x^2))#

Now, use the simplification rule:

#color(white)=color(red)(7x)*color(blue)sqrt(150)*color(blue)(sqrt(x^2))#

#=color(red)(7x)*color(blue)sqrt(150)*color(red)x#

#=color(red)(7x)*color(red)x*color(blue)sqrt(150)#

#=color(red)(7x^2)*color(blue)sqrt(150)#

Technically, this answer is correct, but it can be simplified further by factoring #150# and then using the simplification rule backward again:

#color(white)=color(red)(7x^2)*color(blue)sqrt(150)#

#=color(red)(7x^2)*color(blue)sqrt(6*25)#

#=color(red)(7x^2)*color(blue)sqrt6*color(blue)sqrt25#

#=color(red)(7x^2)*color(blue)sqrt6*color(blue)sqrt(5^2)#

#=color(red)(7x^2)*color(blue)sqrt6*color(red)5#

#=color(red)(7x^2)*color(red)5*color(blue)sqrt6#

#=color(red)(35x^2)*color(blue)sqrt6#

This answer is fully simplified.