How to get rid of the square roots of the denominator: $\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}$?
- I squared the whole denominator, but that didn't help.
- Also I searched for a propriety or identity like $A^2-B^2$, but I didn't see one that could fit.
Any help is appreciated.
$\endgroup$3 Answers
$\begingroup$The trick here is to multiply and divide by the conjugate.
For instance $$\frac{1}{1+\sqrt{2}} = \frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{1-\sqrt{2}}{-1}$$
We can do the same here, but we will do it twice.
$$\frac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{((\sqrt{7}-2\sqrt{5})+\sqrt{3})((\sqrt{7}-2\sqrt{5})-\sqrt{3})} $$
$$=\frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3} = \frac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{30-4\sqrt{35}}$$
Now multiply and divide by $30+4\sqrt{35}$ and you will elliminate the radical here as well.
$\endgroup$ 4 $\begingroup$$$\begin{align}\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}&=\frac{1}{\sqrt 7-2\sqrt 5+\sqrt 3}\cdot \frac{\sqrt 7-2\sqrt 5-\sqrt 3}{\sqrt 7-2\sqrt 5-\sqrt 3}\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{(\sqrt 7-2\sqrt 5)^2-3}\\&=\frac{\sqrt 7-2\sqrt 5-\sqrt 3}{4(6-\sqrt{35})}\cdot\frac{6+\sqrt{35}}{6+\sqrt{35}}\\&=\frac{(\sqrt 7-2\sqrt 5-\sqrt 3)(6+\sqrt{35})}{4}\end{align}$$
$\endgroup$ 0 $\begingroup$To get rid of the square roots of the denominator, you may use $a^2-b^2=(a+b)(a-b)$.
$$\dfrac{1}{\sqrt{7}-2\sqrt{5}+\sqrt{3}}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{(\sqrt{7}-2\sqrt{5})^2-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{7-4\sqrt{35}+20-3}\\=\dfrac{\sqrt{7}-2\sqrt{5}-\sqrt{3}}{24-4\sqrt{35}}\\=\dfrac{(\sqrt{7}-2\sqrt{5}-\sqrt{3})(24+4\sqrt{35})}{24^2-16*35}\\=\dfrac{(\sqrt{7}-2\sqrt{5}-\sqrt{3})(24+4\sqrt{35})}{16}\\=\dfrac{-24\sqrt3-20\sqrt5-16\sqrt7-4\sqrt{105}}{16}\\=\dfrac{-6\sqrt3-5\sqrt5-4\sqrt7-\sqrt{105}}{4}$$
$\endgroup$
