## Precalculus (6th Edition)

$\color{blue}{-\dfrac{\sqrt{21}+7}{4}}$
Rationalize the denominator by multiplying the conjugate of the denominator, which is $\sqrt3+\sqrt7$, to both the numerator and the denominator to obtain: $=\dfrac{\sqrt7(\sqrt3+\sqrt7)}{(\sqrt3-\sqrt7)(\sqrt3+\sqrt7)}$ Distribute $\sqrt7$ in the numerator and use the formula $(a-b)(a+b)=a^2-b^2$ to simplify the denominator to obtain: $=\dfrac{\sqrt{21}+\sqrt{49}}{(\sqrt3)^2-(\sqrt7)^2} \\=\dfrac{\sqrt{21}+7}{3-7} \\=\dfrac{\sqrt{21}+7}{-4} \\=\color{blue}{-\dfrac{\sqrt{21}+7}{4}}$