# solve this algebraic equation.

• Sep 16th 2008, 08:14 PM
paulrb
solve this algebraic equation.
The original problem has to do with the interest rate at which 2 choices have the same present value, but I figured out the equation on my own. For some reason, I can't figure out how to solve it, even though I'm usually very comfortable with solving algebraic equations...

6000 + (5940)/(1+i) =12000/[(1+i)^(1/2)]

The answer is i = .21 (answer in the back of the book, works when plugged in) but I can't figure out how to get it.
• Sep 16th 2008, 08:31 PM
galalio
i was tryinhg to do this homework and i could not do it and it took me so much time.
If f(x) = sin x and f(a) = 2/6, then what is the value of f(a) + f(pi + a) + f(2pi + a)?
• Sep 16th 2008, 08:37 PM
11rdc11
Quote:

Originally Posted by galalio
If f(x) = sin x and f(a) = 2/6, then what is the value of f(a) + f(pi + a) + f(2pi + a)?

• Sep 16th 2008, 08:44 PM
11rdc11
$\frac{6000(1 + i) +5940}{1+i} = \frac{12000}{\sqrt{1+i}}$

$6000(1 + i) +5940 = 12000\sqrt{1+i}$

$6000(1 + i) -12000\sqrt{1+i} +5940 = 0$

let $u = \sqrt{1+i}$

so

$6000u^2 - 12000u + 5940 = 0$

Use quad formula and then remember to sub back in.