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The 2n and n is being raised as a power.

2006-12-16 09:16:26 · 4 answers · asked by Anonymous in Science & Mathematics Mathematics

4 answers

let x^n be u
u^2-4u-5=0
(u-5)(u+1)=0
u=5 0r 1
if u=5 x^n=5 and x=5^(1/n)
if u=1 x^n=1 and x= 1

2006-12-16 09:21:21 · answer #1 · answered by raj 7 · 0 0

x^2n - 4x^n - 5

Don't be confused by the n in the exponent

note that (x^n)^2 = x^nx^n = x^(n + n) = x^2n

You could set x^n = z

z^2 - 4z - 5

(z - 5)(z + 1)

(x^n - 5)(x^n + 1)

2006-12-16 17:25:30 · answer #2 · answered by kindricko 7 · 0 0

Let y = x^n. Then you have:

y^2 - 4y - 5 = (y - 5)(y + 1) = (x^n - 5)(x^n + 1)

2006-12-16 18:22:00 · answer #3 · answered by Northstar 7 · 0 0

x^2n-4x^n-5

(x^n-5)(x^n+1)-final answer

2006-12-16 17:28:53 · answer #4 · answered by ARLONE E 1 · 0 0

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