How do you multiply #sqrt(2x^6) *sqrt(6x^4)#?

Question

2169 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-13T20:18:34

It is

#sqrt(2x^6)*sqrt(6x^4)=sqrt(12x^10)=2*sqrt3*absx^5#

Answer 2 · 2018-03-13T20:31:32

#sqrt(2*x^6) * sqrt(6*x^4) = 3.464*x^5#

There are 2 basic approaches available.

1st Approach
Do the square roots first.
#sqrt(2*x^6) * sqrt(6*x^4) = (1.414*x^3) * (2.449*x^2)#

#sqrt(2*x^6) * sqrt(6*x^4) = 3.464*x^(3+2) = 3.464*x^5#

2nd Approach
Combine the 2 radicals first.
#sqrt(2*x^6) * sqrt(6*x^4) = sqrt(2*6*x^(6+4)#

#sqrt(2*x^6) * sqrt(6*x^4) = sqrt(12*x^10) = 3.464*x^5#

I hope this helps,
Steve

Answer 3 · 2018-03-13T20:37:44

With any problem involving the multiplication of two square roots you can use the property #sqrtx times sqrty = sqrt(xy)#

Since a square root is equivalent to #x^{1/2}# we can use the exponential laws for square roots.
#a^xb^x=(ab)^x# --> #sqrtx times sqrty = sqrt(xy)#

#sqrt(2x^6) times sqrt(6x^4)=sqrt(2x^6 times 6x^4) = sqrt(12x^(10))#

#sqrt(12x^10)# can be simplified to #x^5 times sqrt(12)# since #sqrt(x^10) = x^5#

You could also break out a 2 by dividing the inside by 4 (since #sqrt4 =2#) giving you #2x^5sqrt3#