How do you simplify #10 sqrt(10a) * root4(200a^5)#?

1 Answer
Sep 20, 2015

Answer:

#100*2^(frac(1)(4))*a^(frac(7)(4))#

Explanation:

This is fairly easy to do if you remember this rule (definition):
#sqrt(x) = x^(frac(1)(2))#
#root(3)(x) = x^(frac(1)(3))#
#root(4)(x) = x^(frac(1)(4))#
...
#root(n)(x) = x^(frac(1)(n))#

And the rule:
#(a*b)^x = a^x * b^x#

And the rule:
#a^x * a^y = a^(x+y)#

So let's attack this first by rewriting all those roots:
We have:
#10 sqrt(10a) * root(4)(200a^5)#
which becomes
#=10*(10a)^(frac(1)(2)) * (200a^5)^(frac(1)(4))#
#=10*10^(frac(1)(2)) * a^(frac(1)(2)) * 200^(frac(1)(4)) * (a^5)^(frac(1)(4))#

Now that we have everything in exponential form, let's add apples together and oranges together, i.e., "10" goes together, and "a" goes together.
Let's do it bit by bit.
First, the "10"s.
We have: #10*10^(frac(1)(2))#
This is really saying: #10^1 * 10^(frac(1)(2))#
Using the rule above, we get #10^(1+frac(1)(2)) = 10^(frac(3)(2))#.

Next, we do the "a"s.
We have: #a^(frac(1)(2)) * (a^5)^(frac(1)(4))#
Using the rule above, we get: #a^(frac(1)(2)) * a^(frac(5)(4))#
This becomes: #a^(frac(1)(2) + frac(5)(4)) = a^(frac(2)(4)+frac(5)(4))= a^(frac(7)(4))#

Finally, this #200^(frac(1)(4))#.
Not quite easy, but perhaps rewriting this as square-root of a square-root might do the trick.
So:
#200^(frac(1)(4)) = (200^(frac(1)(2)))^(frac(1)(2))#
Then, this is equal to:
#= ((2*100)^(frac(1)(2)))^(frac(1)(2))#
which becomes
#=(10sqrt(2))^(frac(1)(2))#
#=sqrt(10)*sqrt(sqrt(2))#
or
#=10^(frac(1)(2))*2^(frac(1)(4))#

Putting all the bits together we have:
#=10^(frac(3)(2)) * a^(frac(7)(4)) * 10^(frac(1)(2))*2^(frac(1)(4))#
Putting the "10"s together, we end up having:

#=100*2^(frac(1)(4))*a^(frac(7)(4))#