How do you rationalize the denominator and simplify $\frac{4}{5 + \sqrt{2}}$ $4/(5+sqrt2)$ ?

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How do you rationalize the denominator and simplify $\frac{4}{5 + \sqrt{2}}$ $4/(5+sqrt2)$ ?

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Answer 1 · 2016-03-19T14:09:22

$= \frac{20 - 4 \sqrt{2}}{23}$ $=(20-4sqrt(2))/(23)$

Explanation:

Consider $a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

Multiply by 1 but where $1 = \frac{5 - \sqrt{2}}{5 - \sqrt{2}}$ $1=(5-sqrt(2))/(5-sqrt(2))$

$\frac{4}{5 + \sqrt{2}} \times \frac{5 - \sqrt{2}}{5 - \sqrt{2}} \to \frac{20 - 4 \sqrt{2}}{5^{2} - 2}$ $color(brown)(4/(5+sqrt(2))xx(5-sqrt(2))/(5-sqrt(2)))color(blue)(->(20-4sqrt(2))/(5^2-2)$

$= \frac{20 - 4 \sqrt{2}}{23}$ $=(20-4sqrt(2))/(23)$

Answer 2 · 2016-03-19T14:16:05

The simplified solution is $\frac{4 (5 - \sqrt{2})}{23}$ $(4(5 - sqrt2))/23$

Explanation:

To rationalize the denominator you need to multiply it by a term that will eliminate the surd
You must then multiply the numerator by the same term to keep the value of the fraction the same.
So multiply by $\frac{5 - \sqrt{2}}{5 - \sqrt{2}}$ $(5 - sqrt2)/(5 - sqrt 2)$
this gives

$\frac{4 (5 - \sqrt{2})}{(5 + \sqrt{2}) (5 - \sqrt{2})}$ $(4(5 - sqrt2))/( (5 + sqrt2)(5 - sqrt2)$ = $\frac{4 (5 - \sqrt{2})}{25 - 2}$ $(4(5 - sqrt2))/(25 -2)$

So the solution is $\frac{4 (5 - \sqrt{2})}{23}$ $(4(5 - sqrt2))/ 23$

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