How do you rationalize the denominator and simplify $\frac{\sqrt{15}}{\sqrt{35}}$ $sqrt15/sqrt35$ ?

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How do you rationalize the denominator and simplify $\frac{\sqrt{15}}{\sqrt{35}}$ $sqrt15/sqrt35$ ?

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Answer 1 · 2016-04-07T09:32:24

Very slightly different approach

$\frac{\pm \sqrt{21}}{7}$ $(+-sqrt(21))/7$

Explanation:

Given: $\frac{\sqrt{15}}{\sqrt{35}}$ $" "sqrt(15)/sqrt(35)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the example: $\sqrt{16} = 4$ $sqrt(16) =4$
(actually it is $\pm 4$ $+-4$ )

Write as $\sqrt{4} \times \sqrt{4} = 2 \times 2 = 4$ $sqrt(4)xxsqrt(4) = 2xx2 =4$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the principle demonstrated in the example:

Write as: $\frac{\sqrt{3} \times \sqrt{5}}{\sqrt{5} \times \sqrt{7}}$ $" "(sqrt(3)xxsqrt(5))/(sqrt(5)xxsqrt(7))$

$\Rightarrow \frac{\sqrt{3}}{\sqrt{7}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{3}}{\sqrt{7}}$ $=> sqrt(3)/sqrt(7)xxsqrt(5)/sqrt(5)" " =" " sqrt(3)/sqrt(7)$

Multiply by 1 but in the form of $1 = \frac{\sqrt{7}}{\sqrt{7}}$ $1=sqrt(7)/sqrt(7)$

$\Rightarrow \frac{\sqrt{3}}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}$ $=> sqrt(3)/sqrt(7)xxsqrt(7)/sqrt(7)$

Both 3 and 7 are prime so we have

$\frac{\pm \sqrt{3 \times 7}}{7} = \frac{\pm \sqrt{21}}{7}$ $(+-sqrt(3xx7))/7 =(+-sqrt(21))/7$

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How do you rationalize the denominator and simplify $\frac{\sqrt{15}}{\sqrt{35}}$ $sqrt15/sqrt35$ ?

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How do you rationalize the denominator and simplify √15√35sqrt15/sqrt35?

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How do you rationalize the denominator and simplify $\frac{\sqrt{15}}{\sqrt{35}}$ $sqrt15/sqrt35$ ?