How do you rationalize the dominator and simplify for the square root of 35 over the square root of 55?

Question

How do you rationalize the dominator and simplify for the square root of 35 over the square root of 55?

1 Answer

Impact of this question

2739 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-05T00:52:39

$sqrt(77)/11$

Explanation:

Your expression is $sqrt(35)/sqrt(55)$ . Since you can always mutiply an expression by $1$ without changing its value, and you can see $1$ as any number divided by itself, we have

$sqrt(35)/sqrt(55)=sqrt(35)/sqrt(55)*1 = sqrt(35)/sqrt(55)*sqrt(55)/sqrt(55)$

Doing the multiplications (remember that $sqrt(a)*sqrt(b)=sqrt(a*b)$ ), we have

$sqrt(35*55)/sqrt(55^2)$

And of course $sqrt(a^2)=a$ (if $a$ is positive), so we have

$sqrt(1925)/55$ .

Finally, we can factor $1925$ with prime numbers, and we have

$1925 = 5^2*7*11$ , and so $sqrt(1925)=sqrt(5^2*7*11)=5sqrt(7*11)$ , and $5$ and $55$ cancel out.

Science

Math

Humanities

... and beyond

How do you rationalize the dominator and simplify for the square root of 35 over the square root of 55?

1 Answer

Explanation:

Related questions

Impact of this question