How do you rationalize (2sqrt5-8)/ (2sqrt5+3)?

Mar 4, 2018

$2 \left(2 - \sqrt{5}\right)$

Explanation:

$\frac{2 \sqrt{5} - 8}{2 \sqrt{5} + 3}$. Multiplying by $\left(2 \sqrt{5} - 3\right)$ on

both numerator and denominator we get,

$= \frac{\left(2 \sqrt{5} - 8\right) \left(2 \sqrt{5} - 3\right)}{\left(2 \sqrt{5} + 3\right) \left(2 \sqrt{5} - 3\right)}$

$= \frac{20 - 2 \sqrt{5} \left(8 + 3\right) + 24}{{\left(2 \sqrt{5}\right)}^{2} - {3}^{2}}$

$= \frac{44 - 22 \sqrt{5}}{20 - 9} = \frac{22 \left(2 - \sqrt{5}\right)}{11}$

$= 2 \left(2 - \sqrt{5}\right)$ [Ans]

Mar 4, 2018

$\frac{2 \sqrt{5} - 8}{2 \sqrt{5} + 3} = 4 - 2 \sqrt{5}$

Explanation:

To rationalize the denominator, we multiply by the conjugate and use the difference of squares rule. In this case, the conjugate is $2 \sqrt{5} - 3$, so we multiply by it on both top and bottom:

$\frac{2 \sqrt{5} - 8}{2 \sqrt{5} + 3} = \frac{\left(2 \sqrt{5} - 8\right) \left(2 \sqrt{5} - 3\right)}{\left(2 \sqrt{5} + 3\right) \left(2 \sqrt{5} - 3\right)}$

The difference of squares rule says:
$\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$

Applying this to the denominator, we get:
$\frac{\left(2 \sqrt{5} - 8\right) \left(2 \sqrt{5} - 3\right)}{4 \cdot 5 - 3}$

Then we multiply out the top:
$\frac{20 - 6 \sqrt{5} - 16 \sqrt{5} + 24}{11} = \frac{44 - 22 \sqrt{5}}{11} = 4 - 2 \sqrt{5}$