Simplify #11xxsqrt((49a^5)/(4a^3)#?

2 Answers
Apr 6, 2017

Answer:

Solution 2 of 2

Rather than use shortcuts I have given a lot of detail. This is so that you can see where some of the shortcuts come from.

Simplification is #(77a)/2#

Explanation:

You are looking for squared values/variables that can be 'taken outside' the square root.

Demonstrating a property by example:

Suppose we had #sqrt(a/b)# this can be written as #sqrt(a)/sqrt(b)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:#" "11sqrt((49a^5)/(4a^3)#

Write as :#" "11sqrt(49a^5)/sqrt(4a^3)#

#color(green)("For now forget about the 11. Deal with it at the end.")#

Write as:# " "sqrt(7^2xxa^2xxa^2xxa)/sqrt(2^2xxa^2xxa)#

You can cancel some out at this stage. I will do it later.

'Extracting from the root' we have:

#(7xxaxxa)/(2xxa)xxsqrt(a)/sqrt(a)#

This is the same as:

#7/2xxa/axxaxxsqrt(a)/sqrt(a)#

But #a/a=1 and sqrt(a)/sqrt(a)=1# giving

#7/2xx1xxaxx1 color(green)(larr" turning into 1 is the same as cancelling out")#

#(7a)/2#
Now we deal with the 11: #->11xx(7a)/2=(77a)/2#

Apr 6, 2017

Answer:

Solution 1 of 2: using shortcuts
Jumping steps in my head using the principles shown in solution 2 of 2

#11xx(7a)/2=(77a)/2#

Explanation:

Given:#" "11xxsqrt((49a^5)/(4a^3)#

#11xxsqrt((49a^(5-3))/(4)#

#(77a)/2#