How do you simplify #sqrt735/sqrt5#?
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#sqrt735/sqrt5#
= #sqrt(3xx5xx7xx7)/sqrt5#
= #sqrt((3xx5xx7xx7)/5)#
= #sqrt((3xxcancel5xx7xx7)/cancel5)#
= #sqrt(3xxul(7xx7))#
= #7sqrt3#
#color(blue)("Method")#
We need to look for integer factors of 735 and with a bit of luck be able to cancel out the #sqrt(5)# denominator.
Suppose we had 2 unknown variables #a" and "b#. Suppose these 2 variables were presented in the form of
#sqrt(a)/sqrt(b)# This we can write as #sqrt(a/b)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering your question")#
Notice that the sum of the digits in 735 is 15. As 15 is divisible by 3 then 735 is also divisible by 3
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From the factor tree we observe that 735 is the product of #3xx5xx7^2#
Write #sqrt(735)/sqrt(5)# as #sqrt((3xxcancel(5)xx7^2)/cancel(5))#
We can take the #7^2# outside the square root but it becomes just #7# giving:
#7sqrt(3)#
