Hi could someone please help me solve the following eqation: 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
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Originally Posted by 1234567 Hi could someone please help me solve the following eqation: 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.
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.
Originally Posted by 1234567 Hi could someone please help me solve the following eqation: 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.
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?
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