Question

rationalise the denominator 5/root3-root5

Answer 1

Answer:

The rationalize form is $\frac{5}{\sqrt{3}-\sqrt{5}}=-\frac{5(\sqrt{3}+\sqrt{5})}{2}$

Step-by-step explanation:

Given : Expression $\frac{5}{\sqrt{3}-\sqrt{5}}$

To find : Rationalize the expression ?

Solution :

Expression $\frac{5}{\sqrt{3}-\sqrt{5}}$

Multiply the numerator and denominator by $\sqrt{3}+\sqrt{5}$

$=\frac{5}{\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}$

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

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

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

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

Therefore, The rationalize form is $\frac{5}{\sqrt{3}-\sqrt{5}}=-\frac{5(\sqrt{3}+\sqrt{5})}{2}$

Answer 2

Answer:

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

Step-by-step explanation:Rationalizing the denominator of $\frac{5}{\sqrt{3}-\sqrt{5}}\\multiply\: numerator\: and\; denominator\: by\: \sqrt{3}+\sqrt{5} \:, we\: get\\=\frac{(\sqrt{3}+\sqrt{5})}{(\sqrt{3}-\sqrt{5})(\sqrt{3}+\sqrt{5})}\\=\frac{\sqrt{3}+\sqrt{5}}{(\sqrt{3})^{2} - (\sqrt{5})^{2}}\\=\frac{\sqrt{3}+\sqrt{5}}{(3-5)}\\=\frac{\sqrt{3}+\sqrt{5}}{(-2)} \* denominator\: rationalized *\\\Therefore,\\\frac{5}{(\sqrt{3}-\sqrt{5})}=\frac{-(\sqrt{3}+\sqrt{5})}{2}$