# Imaginary Numbers

• Apr 13th 2009, 12:57 PM
Hapa
Imaginary Numbers
Having trouble with the following problem:

(2 + 2i) (3 - 4i)

I have worked out this much so far.

6 - 8i + 6i - 8i^2

6 - 2i - 8i^2

Since there is an "i" squared, do i distribute that with the preceding "8" coefficient?

Also running into a problem solving the following word problem:

"The sum of the squares of two consecutive positive odd integers is 514. What are the integers?"

I have it set up as x^2 + (x+2)^2 = 514

x^2 + x^2 + 4x + 4 = 514

2x^2 + 4x = 510

Thanks again for everyone's help!
• Apr 13th 2009, 01:15 PM
Reckoner
Quote:

Originally Posted by Hapa
Since there is an "i" squared, do i distribute that with the preceding "8" coefficient?

By definition, $\displaystyle i^2=-1.$ Substitute this into your expression.

Quote:

"The sum of the squares of two consecutive positive odd integers is 514. What are the integers?"

I have it set up as x^2 + (x+2)^2 = 514

x^2 + x^2 + 4x + 4 = 514

2x^2 + 4x = 510
You should be able to factor this:

$\displaystyle 2x^2+4x-510=0$

$\displaystyle \Rightarrow x^2+2x-255=0$

Can you break -255 into two factors whose sum is 2? Start by listing the possible factorizations of -255:

$\displaystyle -255\cdot1\qquad255\cdot-1$

$\displaystyle -17\cdot15\qquad17\cdot-15$

$\displaystyle -5\cdot51\qquad5\cdot-51$

$\displaystyle -3\cdot85\qquad3\cdot-85$

Now which pair adds up to +2?