Question

HELP THIS IS DUE TOMORROW!
Algebra 2 question!
Picture attached below

Answer 1

Answer:

$\dfrac{4+\sqrt{3}}{4}-\dfrac{(3+4\sqrt{3})}{12}i$

Step-by-step explanation:

Given expression:

$\dfrac{4+\sqrt{3}}{3+\sqrt{-3}}$

$\textsf{Apply radical rule}:\quad \sqrt{-a}=\sqrt{a}\sqrt{-1}$

$\implies \dfrac{4+\sqrt{3}}{3+\sqrt{3}\sqrt{-1}}$

$\textsf{Apply imaginary number rule}:\quad \sqrt{-1}=i$

$\implies \dfrac{4+\sqrt{3}}{3+\sqrt{3}i}$

Multiply by the conjugate of the denominator:

$\implies \dfrac{4+\sqrt{3}}{3+\sqrt{3}i} \cdot \dfrac{3-\sqrt{3}i}{3-\sqrt{3}i}$

$\implies \dfrac{(4+\sqrt{3})(3-\sqrt{3}i)}{(3+\sqrt{3}i)(3-\sqrt{3}i)}$

$\implies \dfrac{12-4\sqrt{3}i+3\sqrt{3}-\sqrt{3}i\sqrt{3}}{9-3\sqrt{3}i+3\sqrt{3}i-\sqrt{3}i\sqrt{3}i}$

$\implies \dfrac{12+3\sqrt{3}-4\sqrt{3}i-3i}{9-3i^2}$

$\implies \dfrac{12+3\sqrt{3}-(3+4\sqrt{3})i}{9-3i^2}$

$\textsf{Apply imaginary number rule}:\quad i^2=-1$

$\implies \dfrac{12+3\sqrt{3}-(3+4\sqrt{3})i}{9-3(-1)}$

$\implies \dfrac{12+3\sqrt{3}-(3+4\sqrt{3})i}{12}$

Separate the fractions:

$\implies \dfrac{12+3\sqrt{3}}{12}-\dfrac{(3+4\sqrt{3})i}{12}$

Simplify:

$\implies \dfrac{4+\sqrt{3}}{4}-\dfrac{(3+4\sqrt{3})}{12}i$