How do you simplify #root4(x^12/y^4)#?

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Answer 1 · 2017-02-25T05:55:15

#root4(x^12/y^4) = x^3/y#

#root4(x^12/y^4) = (x^12/y^4)^(1/4) =(x^12 * y^-4)^(1/4)#

Remember the rule of indices: #(a^m)^n = a^(m xx n)#

Applying this rule to the expression:

#(x^12 * y^-4)^(1/4) = x^(12/4) * y^(-4/4)#

#= x^3 * y^-1#

#= x^3/y#

Answer 2 · 2017-02-25T06:00:14

Detailed explanation:

#color(blue)(root(4)((x^12)/(y^4))#

Let's solve it using simple steps

First we should know that #color(brown)(root(x)(y/z)=(root(x)(y))/(root(x)(z))#

So,

#rarrroot(4)((x^12)/(y^4))=(root(4)(x^12))/(root(4)(y^4))#

Now solve #root(4)(x^12)# and #root(4)(y^4)# each independantly

Let's solve #root(4)(x^12)# (expand it)

#rarrroot(4)(x^12)=root(4)(x*x*x*x*x*x*x*x*x*x*x*x)#

Take out the roots

#rarrroot(4)(underbrace(x*x*x*x)*underbrace(x*x*x*x)*underbrace(x*x*x*x))#

#rarrx*x*x#

#color(green)(rArrx^3#

Now solve #root(4)(y^4)# (expand it)

#rarrroot(4)(y^4)=root(4)(y*y*y*y)#

Take out the roots

#rarrroot(4)(underbrace(y*y*y*y))#

#color(green)(rArry#

So, put the solutions together to get our answer

#color(blue)(x^3/y#

Hope this helps! :)