Two algebra ll problems I need help with

1) If i = the square root of negative one, what is one over a + bi?

2) There are two numbers with the following properties:
The second number is 3 more than the first number
The product of the two numbers is 9 more than their sum


Any help is appreciated

You phrased the first question a little awkward, but I can answer the second. :3

A = first number, B = second number, C = the sum of A and B

B - A = 3
A + B = C; (A x B) - C = 9

The only logical numbers that work this way are if:
A = 3
and
B = 6

That way,
(6) - (3) = 3
(3) + (6) = 9; ( (3) x (6) ) - 9 = 9.

daSVgrouch: why not -4 and -1 also?
-4 * -1 = 4
-4 + -1 = -5

-5 + 9 = 4 works as well...

***
and clarification on #1:
1 / (a + bi) = (a - bi) / [(a + bi)(a - bi)]
= (a - bi) / [a^2 - b^2 i^2]
but i^2 = -1, so this becomes:
1 / (a + bi) = (a - bi) / [a^2 + b^2]

the general technique for dividing complex numbers is to multiply top and bottom by the conjugate of the bottom (the conjugate changes the sign of the imaginary part: the conjugate of a + bi is a - bi)

1. the inverse

2. let first number = a
second number = (a + 3)
product of (a)(a + 3a) = 9 + a + a + 3

a^2 + 3a = 2a + 12

a^2 + a - 12 = 0

(a - 4)(a + 3)

1) 1/(a+ib) = (a-ib)/(a+ib)*(a-ib) = (a-ib)/ (a^2+b^2) = {a/(a^2+b^2)- i{b/(a^2+b^2)}.....Ans

2) b-a = 3.....(i) where b> a
ab = a+b+9..........(ii)
from (i) b = a+3..............(iii)
from (ii)
a(a+3) = a+(a+3) +9
a^2 +3a = 2a +12
a^2 +a -12 =0
(a+4)(a-3) =0
a = 3 or a= -4.............(iii)
Therefore b = 3+3=6 or b = -4+3 = -1...............(iv)
Therefore two numbers are (3,6) or (-4,-1).............Ans

1)
1 / (a+bi) = ( a - b i) / (a^2 + b^2)

2) (x+3)x = 9+2x+3
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3
x+3 = 6