How do you simplify $\frac{1}{3 + \sqrt{5}}$ $1/(3+sqrt5)$ ?

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Answer 1 · 2018-04-07T13:29:05

$\frac{3 - \sqrt{5}}{4}$ $(3-sqrt5)/4$

Rationalize the denominator (radicals cannot be in the denominator) by multiplying the numerator and denominator by $3 - \sqrt{5}$ $3-sqrt5$

$\frac{1}{3 + \sqrt{5}}$ $1/(3+sqrt5)$

$\frac{1 \cdot 3 - \sqrt{5}}{(3 + \sqrt{5}) (3 - \sqrt{5})}$ $(1*3-sqrt5)/((3+sqrt5)(3-sqrt5))$

$(3 + \sqrt{5}) (3 - \sqrt{5}) = 9 - 3 \sqrt{5} + 3 \sqrt{5} - \sqrt{25}$ $(3+sqrt5)(3-sqrt5)=9-3sqrt5+3sqrt5-sqrt25$

$9 - 3 \sqrt{5} + 3 \sqrt{5} - \sqrt{25} = 9 - 5 = 4$ $9-3sqrt5+3sqrt5-sqrt25=9-5=4$

$\frac{3 - \sqrt{5}}{4}$ $(3-sqrt5)/4$

Answer 2 · 2018-04-07T13:30:29

$\frac{3 - \sqrt{5}}{4}$ $(3-sqrt(5))/4$

You can rationalise the denominator by multiplying by it’s conjugate.

$\frac{1}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}}$ $1/(3+sqrt(5))times(3-sqrt(5))/(3-sqrt(5))$

$= \frac{3 - \sqrt{5}}{9 - 5}$ $=(3-sqrt(5))/(9-5)$

$= \frac{3 - \sqrt{5}}{4}$ $=(3-sqrt(5))/4$

How do you simplify 13+√51/(3+sqrt5) ?