# Solving equation

• Nov 27th 2010, 11:52 AM
1234567
Solving equation

4(1+r)^4-2(1+r)^2-3=0

I Know if it was a qudratic, just use the formula, but how does one solve for degree 4.

Thanks
• Nov 27th 2010, 12:00 PM
skeeter
Quote:

Originally Posted by 1234567

4(1+r)^4-2(1+r)^2-3=0

I Know if it was a qudratic, just use the formula, but how does one solve for degree 4.

Thanks

let $t = (1+r)^2$

$4t^2 - 2t - 3 = 0$

use the quadratic formula and solve for t (remember that t will have to be > 0 ) ... then solve for r.
• Nov 27th 2010, 12:05 PM
1234567
Quote:

Originally Posted by skeeter
let $t = (1+r)^2$

$4t^2 - 2t - 3 = 0$

use the quadratic formula and solve for t (remember that t will have to be > 0 ) ... then solve for r.

Thanks for the hint, i should manage to solve the problem now.
• Nov 27th 2010, 12:06 PM
Quote:

Originally Posted by 1234567

4(1+r)^4-2(1+r)^2-3=0

I Know if it was a qudratic, just use the formula, but how does one solve for degree 4.

Thanks

You could substitute

$x=(1+r)^2$

$4x^2-2x-3=0$

Now use the quadratic formula, realising that x is a square, hence positive,
so you only need to take one solution for x.

Rearranging your solution for x will achieve the 2 solutions for r.
• Nov 28th 2010, 01:46 AM
HallsofIvy
Quote:

$x=(1+r)^2$
$4x^2-2x-3=0$