A Nice Nonic Equation | Problem 408

[@Blaqjaqshellaq]

Another thing you can do is consider that (z+1)^10 - z^10=0, therefore [(z+1)^5+z^5]* [(z+1)^5-z^5]=0. If the first factor equals zero, it follows that 2*z^5 + 5*z^4 + 10*z^3 + 10*z^2 + 5*z + 1=0. We can see that if z=-1/2, then (z+1)^5 + z^5=0, so we can factor out (2*z+1), giving us z^4 + 2*z^3 + 4*z^2 + 3*z + 1. On the other hand, if it's the second factor that equals zero, it follows that 5*z^4 + 10*z^3 + 10*z^2 + 5*z + 1=0. It's still pretty complicated, but a pair of quartics are less so than a single nonic.

[@the_m_original]

z = -0.5 solution: (z+1)^10 = z^10 both exponents are divisible by 2, therefore neither, both or one of the bases can be negative. check neither: z+1 = z no solution check both: -z-1 = -z no solution check left: -z-1 = z -2z = 1 z = -0.5 check right: z+1 = -z 2z = -1 z = -0.5 answer is z = -0.5

[@MAZHOR1113]

What's with your mic?

[@rodrigocarbia7386]

Question Can’t you just fifth root both sides and you’re left with a squared binomial that you can just develop which cancels the z squared on the other side and be left with 2z + 1 = 0

[@oliviacercel5742]

At 7:17 the exponent of "e" is doubled. Why?! 🤔

[@RuleofThehyperbolic]

interesting equation: find all solutions for z in terms of x x^(a+z) = x^a

[@the_m_original]

(x^a)(x^z) = x^a x^z = 1 e^zlnx = e^i2PIn zlnx = i2PIn z = (i2PIn)/lnx for any integer n. only real solution for z is 0 where n = 0

[@zzz-lo8vg]

When you get your real solution for z, wouldn't you be able to solve for the complex solutions by multiplying by 10th root of unity

[@scottleung9587]

Cool!

[@fibroidss1194]

But it’s a decic not a nonic

[@Etienne-pq3dx]

When you develop the expression on the left hand side, you get z^10 on both sides, so the highest power is 9 (nonic)

[@mcwulf25]

I did what you did although I put (z+1)/z = t Where t is a 10th root of unity. (1-t)z + 1 = 0 z = 1/(t-1) The two real values of t are 1 and -1. t=1 is undefined. t=-1 gives z = -1/2. I didn’t bother working ot the complex solutions but they will be where t = e^pi.n/5.

[@Blaqjaqshellaq]

If t=1, then z is either 1/2 + i*(3/4)^1/2 or 1/2 - i*(3/4)^1/2.

[@neuralwarp]

Same again then?

[@firstnamelastname442]

doesn’t even need to be complex, -0.5 works

[@hacxman1]

yes, it took me like 2 seconds. you dont even need to touch the pen