How do you simplify $sqrt735/sqrt5$ ?

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Answer 1 · 2016-06-30T12:36:37

$sqrt735/sqrt5=7sqrt3$

$sqrt735/sqrt5$

= $sqrt(3xx5xx7xx7)/sqrt5$

= $sqrt((3xx5xx7xx7)/5)$

= $sqrt((3xxcancel5xx7xx7)/cancel5)$

= $sqrt(3xxul(7xx7))$

= $7sqrt3$

Answer 2 · 2016-07-04T10:59:48

$7sqrt(3)$

$color(blue)("Method")$

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
Tony B

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:

$7sqrt(3)$

How do you simplify sqrt735/sqrt5sqrt735/sqrt5?