Rationalize the denominator of $\frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$

Question

I have to rationalize the denominator of $A = \frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$. I multiplied the fraction by $\frac{\sqrt{7\sqrt{3}-4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}$, and $A=\frac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$. Then I multiplied by $\frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}+4\sqrt{5}}$, and $A=\frac{7\sqrt{201}+4\sqrt{335}}{67}$. Can someone tell me if there is a better solution and whether I am right?

Answer 1

Hint: Multiply numerator and denominator by $$\sqrt{7\sqrt{3}+4\sqrt{5}}$$ and then by $$\sqrt{67}$$ For your work we get $$\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67}$$

Answer 2

Various others have suggested mulrltiplying by $\sqrt{7\sqrt{3}\color{blue}{+}4\sqrt{5}}$ in the first step. The reason this is preferable is because you get a product with at most one square root in the denominator, which simplifies the computation; thus

$\dfrac{7\sqrt{3}+4\sqrt{5}}{\color{blue}{\sqrt{67}}}$

versus your expression

$\dfrac{\sqrt{67}}{7\sqrt{3}-4\sqrt{5}}$

The suggested alternative has a hidden advantage. Though not in this case, sometimes the remaining square root has a squared quantity in the denominator allowing a step to be saved. For instance, compare

$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}+4\sqrt{5}}}{\sqrt{4}}=\dfrac{\sqrt{2\sqrt{21}+4\sqrt{5}}}{2}$

with

$\dfrac{1}{\sqrt{2\sqrt{21}-4\sqrt{5}}}=\dfrac{\sqrt{2\sqrt{21}-4\sqrt{5}}}{2\sqrt{21}-4\sqrt{5}}$

Answer 3

For positive $a$ and $b$, we have $\sqrt{a}/\sqrt{b}=\sqrt{a/b}$. So you can concentrate on $$ \frac{7\sqrt{3}+4\sqrt{5}}{7\sqrt{3}-4\sqrt{5}}=\frac{(7\sqrt{3}+4\sqrt{5})^2}{49\cdot3-15\cdot5}=\frac{(7\sqrt{3}+4\sqrt{5})^2}{67} $$ Thus you're done with $$ \frac{\sqrt{7\sqrt{3}+4\sqrt{5}}}{\sqrt{7\sqrt{3}-4\sqrt{5}}}=\frac{(7\sqrt{3}+4\sqrt{5})\sqrt{67}}{67} $$