Question

8.5 divide radical expressions (homework)
score: 11/14 answered: 12/14
question 13
rationalize the denominator:
\\(\frac{\sqrt{z}-\sqrt{3}}{\sqrt{z}+\sqrt{3}}\\)
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Explanation:

Step1: Multiply by conjugate

Multiply the fraction $\frac{\sqrt{z}-\sqrt{3}}{\sqrt{z}+\sqrt{3}}$ by $\frac{\sqrt{z}-\sqrt{3}}{\sqrt{z}-\sqrt{3}}$.
\[
\frac{\sqrt{z}-\sqrt{3}}{\sqrt{z}+\sqrt{3}}\times\frac{\sqrt{z}-\sqrt{3}}{\sqrt{z}-\sqrt{3}}=\frac{(\sqrt{z}-\sqrt{3})^2}{(\sqrt{z}+\sqrt{3})(\sqrt{z}-\sqrt{3})}
\]

Step2: Expand numerator and denominator

Expand the numerator $(\sqrt{z}-\sqrt{3})^2$ using $(a - b)^2=a^2-2ab + b^2$ and denominator using $(a + b)(a - b)=a^2 - b^2$.
The numerator $(\sqrt{z}-\sqrt{3})^2=z-2\sqrt{3z}+ 3$.
The denominator $(\sqrt{z}+\sqrt{3})(\sqrt{z}-\sqrt{3})=z - 3$.
So the fraction becomes $\frac{z-2\sqrt{3z}+3}{z - 3}$.

Answer:

$\frac{z-2\sqrt{3z}+3}{z - 3}$