1. ## 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

2. 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 $\displaystyle t = (1+r)^2$

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

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

3. Originally Posted by skeeter
let $\displaystyle t = (1+r)^2$

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

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.

4. 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

$\displaystyle x=(1+r)^2$

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

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.

5. Originally Posted by Archie Meade
You could substitute

$\displaystyle x=(1+r)^2$

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

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.
Why are you assuming only real solutions are wanted?