Question

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

Answer 1

`(4+√(3))/(4)-((3+4√(3)))/(12)i`

Explanation:

Given expression:

`(4+√(3))/(3+√(-3))`

`\textsf{Apply radical rule}:\quad √(-a)=√(a)√(-1)`

`\implies (4+√(3))/(3+√(3)√(-1))`

`\textsf{Apply imaginary number rule}:\quad √(-1)=i`

`\implies (4+√(3))/(3+√(3)i)`

Multiply by the conjugate of the denominator:

`\implies (4+√(3))/(3+√(3)i) \cdot (3-√(3)i)/(3-√(3)i)`

`\implies ((4+√(3))(3-√(3)i))/((3+√(3)i)(3-√(3)i))`

`\implies (12-4√(3)i+3√(3)-√(3)i√(3))/(9-3√(3)i+3√(3)i-√(3)i√(3)i)`

`\implies (12+3√(3)-4√(3)i-3i)/(9-3i^2)`

`\implies (12+3√(3)-(3+4√(3))i)/(9-3i^2)`

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

`\implies (12+3√(3)-(3+4√(3))i)/(9-3(-1))`

`\implies (12+3√(3)-(3+4√(3))i)/(12)`

Separate the fractions:

`\implies (12+3√(3))/(12)-((3+4√(3))i)/(12)`

Simplify:

`\implies (4+√(3))/(4)-((3+4√(3)))/(12)i`