2 years ago

4/1 sqrt3-sqrt2 rationalize?

1 answer:

8 0

$\dfrac{4}{\sqrt3-\sqrt2}=\dfrac{4}{\sqrt3-\sqrt2}\cdot\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\dfrac{4(\sqrt3+\sqrt2)}{(\sqrt3)^2-(\sqrt2)^2}\\\\=\dfrac{4\sqrt3+4\sqrt2}{3-2}=4\sqrt3+4\sqrt2$

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