Solving A Decic Polynomial Equation
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Since all of the terms have even powers, let y = x² to begin with. Then y² + y³ = 2y⁵, or by rearranging and factoring,
y²(2y³ - y - 1) =0
This immediately gives y² = x⁴ = 0, x = 0.
By inspection, y = 1 is a solution, so x² = 1, x = ±1.
Divide by (y - 1) to get 2y² - 2y - 1 = 0. Solve by quadratic equation to get y = (1 ± 𝒊)/2 = x². Solve that by polar-coordinate stuff which I am too rusty on.
Nice!
It's worth pointing out that since the original equation is a polynomial of degree 10, there should be 10 solutions. In this problem, you have to count repeated roots, but this lets us know we found them all. (For instance, the common factor of x^4 gives us the solution x=0 with a multiplicity of 4.)
Synthetic division gives the cofactors a tad bit faster without having to be clever in your rewriting to use factoring identities.
Buenas tares Sres. SyberMath, Gracias es un buen ejercicio. Éxitos.i
Gracias 😍
I just substituted y for x^2 and got all the non-trivial solutions from there.
Factor out x^2, it's a little easier than dealing with the quintic.
x^2?
😊😊😊👍👍👍
idk man i substitute 1 in and the equation become true. I guess my strat was the fastest then 😅😅😅
x = 1
Your best video so far (and I’ve watched most of them!).
Glad to hear that!
x⁴(2x⁶ - x² - 1) = 0
x⁴ = 0 => *x = 0*
2x⁶ - x² - 1 = 0
x⁶ - x² + x⁶ - 1 = 0
x²(x⁴ - 1) + (x² - 1)(x⁴ + x² + 1) = 0
(x² - 1)(x² + 1)x² + (x² - 1)(x⁴ + x² + 1) = 0
(x² - 1)(x⁴ + x² + x⁴ + x² + 1) = 0
(x² - 1)(2x⁴ + 2x² + 1) = 0
x² - 1 = 0 => *x = ± 1*
2x⁴ + 2x² + 1 = 0
x² = (-2 ± 2i)/4 = (-1 ± i)/2
*x = ± √[(-1 ± i)/2]*
"Wait, you forgot..." You'd think by now I'd be hip to this notion of factoring by grouping, but I guess I'm too old school to see it on my own. I'm like that old horse that knows one way back to the barn, and insists on going that way long after the path has been buried under development. At least I've still got enough brain cells left to be delighted watching you do these things. Keep it up, please.
Thank you!
x ∈ { -1, 0, 1,
-½ √2√(i-1),
-½ i √2√(i+1),
½ √2√(i-1),
½ i √2√(i+1) }
0 is a quadruple root.(x⁴ = 0)
10 roots there are.
x⁴ +x⁶ = 2x¹⁰
x⁴ [(1+x²)-2x⁶] = 0
By ZPP x = 0 is a quadruple root.
1+x²-2x⁶=0
2x⁶-x²-1=0
u = x²
2u³-u-1=0
u=1 is a root
x= -1, 1
2u² +2u+1=0
...
Hello! Could you please come up with a series on Differential Calculus?
Isn't it called "dectic", not "decic"?
en.wikipedia.org/wiki/Degree_of_a_polynomial
@@SyberMath Honestly, "dectic" sounds better
First
Ich habe jetzt mal spontan, bevor ich es mir angesehen habe 1, -1, -i und i. Sollte passen, aber ich lasse mich gern eines besseren belehren.
1 and -1 are right. But i and -i are wrong. And 0 (zero) is also right.
@@SidneiMV okay, I see my fault. Didn't consider that ^4 is 1 and ^6 -1.
x^10 - x^4 + x^10 - x^6 = 0
x^4 [x^6 - 1 + x^6 - x^2] = 0
x^4 *( x^2 - 1)
*(x^4 + x^2 + 1 + x^2 * (x^2 + 1)) = 0
x^4 * ( x^2 - 1) ( x^4 + x^2 + 1/2) = 0
Hereby
x = 0, 1, -1