simplify:

**Mr_Green** · November 7th 2006, 04:49 PM

simplify:

3/i + 6/(i+2) + i/(i-2)

**topsquark** · November 7th 2006, 04:56 PM

Originally Posted by Mr_Green

simplify:

3/i + 6/(i+2) + i/(i-2)

i is a square root. Thus we wish to get rid of the square roots in the denominators. You do this by multiplying the denominator by the complex conjugate of the denominator.

For example:
$\frac{6}{i+2} = \frac{6}{i+2} \cdot \frac{-i + 2}{-i + 2}$

= $\frac{6(-i+2)}{-i^2 + 4} = \frac{12 - 6i}{1+4} = \frac{12-6i}{5}$

Can you take it from here?

-Dan

**Mr_Green** · November 7th 2006, 07:11 PM

Originally Posted by topsquark

i is a square root. Thus we wish to get rid of the square roots in the denominators. You do this by multiplying the denominator by the complex conjugate of the denominator.

For example:
$\frac{6}{i+2} = \frac{6}{i+2} \cdot \frac{-i + 2}{-i + 2}$

= $\frac{6(-i+2)}{-i^2 + 4} = \frac{12 - 6i}{1+4} = \frac{12-6i}{5}$

Can you take it from here?

-Dan

I'm still stuck! Please help

**Jameson** · November 7th 2006, 08:14 PM

Well he did one of the terms. Are you paying attention to his method? The method is the key.

$\frac{3}{i} \times \frac{i}{i}=-3i$

$\frac{i}{i-2} \times \frac{i+2}{i+2}=...$

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