Question #69536

2 Answers
Pallu · EZ as pi
Oct 28, 2017

We get, # 5 xy^3sqrt(6xz)# from #sqrt(150x^3y^6z)#

Explanation:

#5xy^3sqrt(6xz)# from #sqrt(150x^3y^6z)#

Consider
#sqrt(150x^3y^6z)#

As #a^ mxxa^n = a^(m+n)#,

#=> sqrt((25xx6)xx(x^2xxx)xx(y^2xxy^4)xxz)#

#=> sqrt((5^2xx6)xx(x^2xxx )xx(y^2xxy^2xxy^2)xxz)#

Taking the roots of the square terms outside the radical sign:

#=> 5xx(x)xxyxxyxxyxx sqrt(6xz)#

# =>5 xy^3sqrt(6xz)#

Tony B
Oct 28, 2017

#5xy^3sqrt(6xz)#

Explanation:

#color(green)("The trick is to look for squared values")#

#color(blue)("Dealing with the number part")#

We have 150.

Just for a moment forget that this is hundreds and just focus on the 15 part. It is known that #3xx5=15#

Now we deal with the hundreds part. It is known that 15xx10=150. So putting all of this together we have: #3xx5xx10=150#

We know that #2xx5=10#. Again we can substitute this back in giving: #3xx5xx2xx5=150#

Notice that within this we have #5xx5# so we can write:

#3xx2xx5^2#

Bothe 3 and 2 are prime numbers so there is no way we can square root them. However, #sqrt(5^2)=5# so we can square root that part.

#color(red)(sqrt(150)=5sqrt(3xx2)=5sqrt(6))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Dealing with the variable part (letters)")#

#x^3y^6z#

#x^3# is the same as #x^2xx x#
#y^6# is the same as #y^(2+2+2)=y^2xxy^2xxy^2#

A single #z# we can do nothing with.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

So we have: #5sqrt(6xxx^2 xx x xxy^2xxy^2xxy^2xxz)#

Taking all the squared values out of the root

#5xy^3sqrt(6xz)#

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