Before doing anything too complicated, a solution is x=1, because 1 to the power of anything is 1. However, there are probably other solutions too. After fiddling around awhile, it looks like -1 can also be a solution. The rest of it would do my head in :)
x^17-x^9-x^8+1=0 has 17 roots. x=±1 are two of them. There is third real root. What is it? A plot of x^17-x^9-x^8+1=0 will reveal that x=1 is a double root. The remaining 14 roots are complex. x=±i are two of those complex roots.
There’s a much quicker way to find all real solutions without factoring further then the very beginning. When you have x^8-1 and x^9-1 both equal to 0, x^8 is going to look like a parabola but rise faster than x^2 and since it’s shifted down 1 unit it will only have two real solutions. And the nonic equation looks like x^3 but again rises much faster than x^3 and is shifted down one unit and will only have 1 solution since it is always increasing. So therefore there are only 3 real solutions leaving 14 complex solutions. So now you can take the 8th root of 1 and get +or-1 and lose 6 solutions since we don’t care about them in this case. And with the nonic we just take the 9th root of 1 and get our 1 real solution and we are done. No need to factor and use substitution and quadratic formula. This video can be 3 minutes long.
Before doing anything too complicated, a solution is x=1, because 1 to the power of anything is 1. However, there are probably other solutions too. After fiddling around awhile, it looks like -1 can also be a solution.
The rest of it would do my head in :)
x^17-x^9-x^8+1=0 has 17 roots. x=±1 are two of them. There is third real root. What is it? A plot of x^17-x^9-x^8+1=0 will reveal that x=1 is a double root. The remaining 14 roots are complex. x=±i are two of those complex roots.
Just by watching i can say X=1 is a solution
There’s a much quicker way to find all real solutions without factoring further then the very beginning. When you have x^8-1 and x^9-1 both equal to 0, x^8 is going to look like a parabola but rise faster than x^2 and since it’s shifted down 1 unit it will only have two real solutions. And the nonic equation looks like x^3 but again rises much faster than x^3 and is shifted down one unit and will only have 1 solution since it is always increasing. So therefore there are only 3 real solutions leaving 14 complex solutions. So now you can take the 8th root of 1 and get +or-1 and lose 6 solutions since we don’t care about them in this case. And with the nonic we just take the 9th root of 1 and get our 1 real solution and we are done. No need to factor and use substitution and quadratic formula. This video can be 3 minutes long.
Why not just directly discard (x⁴+1) and (x²+1) that are always ≥1>0 when x is real?
X is +1 or -1
Some of the complex solutions can be easily written in radicals. The others are more simply expressed in trigonometric form.
❌ = 1
1, -1, i, -i
Боже, я не знаю. что такое "омега" и "омега^2" !