Question

8.5 divide radical expressions (homework) score: 7/14 answered: 7/14 question 8 rationalize the denominator: $\frac{2}{sqrt{7}+sqrt{3}}=$ question help: video written example submit question jump to answer

Explanation:

Step1: Multiply by conjugate

Multiply the fraction $\frac{2}{\sqrt{7}+\sqrt{3}}$ by $\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}$.
$\frac{2}{\sqrt{7}+\sqrt{3}}\times\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}-\sqrt{3}}$

Step2: Expand numerator and denominator

Use the FOIL method in the numerator ($2(\sqrt{7}-\sqrt{3}) = 2\sqrt{7}-2\sqrt{3}$) and the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$ in the denominator ($(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})=(\sqrt{7})^{2}-(\sqrt{3})^{2}=7 - 3=4$).
$\frac{2(\sqrt{7}-\sqrt{3})}{(\sqrt{7})^{2}-(\sqrt{3})^{2}}=\frac{2\sqrt{7}-2\sqrt{3}}{4}$

Step3: Simplify the fraction

Divide both the numerator and denominator by 2.
$\frac{2\sqrt{7}-2\sqrt{3}}{4}=\frac{\sqrt{7}-\sqrt{3}}{2}$

Answer:

$\frac{\sqrt{7}-\sqrt{3}}{2}$