How do you simplify #sqrt(15n^2)*sqrt(10n^3)#?

2 Answers
Apr 13, 2017

Answer:

#5n^2sqrt(6n)#

Explanation:

Demonstrating a principle by example:

#sqrt(an)=sqrt(a)xxsqrt(n)#

#sqrt(4xx25)=2xx5=10#
#sqrt(4xx25)=sqrt(100)=10#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "sqrt(15n^2)xxsqrt(10n^3)#

#" "[sqrt(3)xxsqrt(5)xxcolor(blue)(sqrt(n^2))]xx[sqrt(2)xxsqrt(5)xxcolor(red)(sqrt(n^2))xxsqrt(n)]#

#larr" "color(blue)(n)[sqrt(3)xxsqrt(5)]" "xx" "color(red)(n)[sqrt(2) xxsqrt(5)xxsqrt(n)]#
#|" "#......................................................................................................
#|#
#|#
#rarr" "color(brown)(n)[sqrt(3)xxcolor(blue)(sqrt(5))]" "xx" "color(brown)(n)[sqrt(2) xxcolor(blue)(sqrt(5))xxsqrt(n)]#

#" "color(blue)(5)color(brown)(n^2)[sqrt(3)xxsqrt(2)xxsqrt(n)] #

#color(white)()#

#" "5n^2sqrt(6n)#

Apr 13, 2017

Answer:

#5n^2sqrt(6n)#

Explanation:

#color(blue)(sqrt(15n^2)*sqrt(10n^3)#

Split the equation using

#color(brown)(sqrt(xy)=sqrtx*sqrty#

So,

#rarrsqrt15*color(red)(sqrt(n^2))*sqrt(10)*color(red)(sqrt(n^2))*n#

#rarrcolor(red)(n*n)*underbracesqrt5*sqrt3*underbracesqrt5*sqrt2*n#

#color(green)(rArr5n^2*sqrt(6n)#

Hope this helps... :)