## Romero 3 years ago State the equivalent form of

1. Romero

$\frac{2}{3+i}$

2. Romero

$a) \frac{3-i}{4}$ $b) \frac{3-i}{5}$ $c)\frac{4-i}{4}$ $d) \frac{4-i}{5}$

3. Romero

I said the answer was a but I got it wrong. It says that it suppose to be b can anyone explain?

4. gogind

$\frac{2}{3+i} = \frac{2}{3+i} \frac{3-i}{3-i} = \frac{2(3-i)}{3^2 -i^2} = \frac{2(3-i)}{9-(-1)} = \frac{3-i}{5}$

5. denbiner

because i is a radical... you cannot have a radical in the denominator so you need to rationalize the denominator so that the radical goes away from the denom..... so you multiply everything times its conjugate and simplify