How do you rationalise the denominator of $\frac{8}{3 - \sqrt{5}}$ $8/(3-sqrt5)$ ?

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How do you rationalise the denominator of $\frac{8}{3 - \sqrt{5}}$ $8/(3-sqrt5)$ ?

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Answer 1 · 2016-04-03T20:03:08

Multiply both numerator and denominator by $3 + \sqrt{5}$ $3+sqrt(5)$ and simplify.

Explanation:

$\frac{8}{3 - \sqrt{5}}$ $8/(3-sqrt(5))$

$= \frac{8 \cdot (3 + \sqrt{5})}{(3 - \sqrt{5}) (3 + \sqrt{5})}$ $=(8*(3+sqrt(5)))/((3-sqrt(5))(3+sqrt(5)))$

$= \frac{24 + 8 \sqrt{5}}{3^{2} - {(\sqrt{5})}^{2}}$ $=(24+8sqrt(5))/(3^2-(sqrt(5))^2)$

$= \frac{24 + 8 \sqrt{5}}{9 - 5}$ $=(24+8sqrt(5))/(9-5)$

$= \frac{24 + 8 \sqrt{5}}{4}$ $=(24+8sqrt(5))/4$

$= 6 + 2 \sqrt{5}$ $=6+2sqrt(5)$

The expression $(3 + \sqrt{5})$ $(3+sqrt(5))$ is called the conjugate of $(3 - \sqrt{5})$ $(3-sqrt(5))$

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How do you rationalise the denominator of $\frac{8}{3 - \sqrt{5}}$ $8/(3-sqrt5)$ ?

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