Solving A Decic Polynomial Equation
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@jpolowin0 24 күн бұрын
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.
@SyberMath 24 күн бұрын
Nice!
@TedHopp 23 күн бұрын
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.)
@DonRedmond-jk6hj 24 күн бұрын
Synthetic division gives the cofactors a tad bit faster without having to be clever in your rewriting to use factoring identities.
@maxwellarregui814 23 күн бұрын
Buenas tares Sres. SyberMath, Gracias es un buen ejercicio. Éxitos.i
@SyberMath 19 күн бұрын
Gracias 😍
@scottleung9587 23 күн бұрын
I just substituted y for x^2 and got all the non-trivial solutions from there.
@phill3986 24 күн бұрын
😊😊😊👍👍👍
@chasmosaurus3 24 күн бұрын
Factor out x^2, it's a little easier than dealing with the quintic.
@SyberMath 24 күн бұрын
x^2?
@SidneiMV 24 күн бұрын
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]*
@DonEnsley-mathdrum 24 күн бұрын
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 ...
@charleskrueger5523 24 күн бұрын
Your best video so far (and I’ve watched most of them!).
@SyberMath 24 күн бұрын
Glad to hear that!
@rakenzarnsworld2 24 күн бұрын
x = 1
@lesnyk255 24 күн бұрын
"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.
@SyberMath 24 күн бұрын
Thank you!
@HATTRICK202 24 күн бұрын
Hello! Could you please come up with a series on Differential Calculus?
@1Otnt4354g 23 күн бұрын
idk man i substitute 1 in and the equation become true. I guess my strat was the fastest then 😅😅😅
@braydentaylor4639 24 күн бұрын
Isn't it called "dectic", not "decic"?
@SyberMath 24 күн бұрын
en.wikipedia.org/wiki/Degree_of_a_polynomial
@braydentaylor4639 24 күн бұрын
@@SyberMath Honestly, "dectic" sounds better
@HATTRICK202 24 күн бұрын
First
@ElvisTB 24 күн бұрын
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.
@SidneiMV 24 күн бұрын
1 and -1 are right. But i and -i are wrong. And 0 (zero) is also right.
@ElvisTB 24 күн бұрын
@@SidneiMV okay, I see my fault. Didn't consider that ^4 is 1 and ^6 -1.
@honestadministrator 24 күн бұрын
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
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