Mathematicians: How would you find the 6 roots (real or not) from x^6+7x^3 - 8=0 usîng an algebraic method?

Question

Im sorry, I posted the wrong equation last time. I ask those who asked it an apology.

Hey, -2 and 1 are the real roots and they arent triple roots, The problem is how to find the complex ones. Theres a very good answer, which is equals to my own one.

I was meaning to find out if someone has a different method to solve this question

Ana

Answer 1

if we put x^3 =t we get quadratic in t
t^2+7t-8=0
(t-8)(t+1) =0
t= 8 or -1

now x^3 =8 or -1
so x = 2,2w,2w^2 , -1,-w,-w^2 where w is cube root of 1
this is well defined however I shall derive the same

x^3+1 =0
=>(x-1)(x^2-x+1) = 0
=>x= 1 or solving x^2-x+1 = 0 we get 2 roots

x = (1+/-sqrt(-3))/2 using x = (=-b+/-sqrt(b^2=2ac)/2a

There is another method for solving x^3=8

x^3 = 2^3.1
= 2^3.e^2*pi*i
= 2^3.e^2*n*pi*i
x =2 e^(2*n*pi*i)/3 for n =0,1,2 usin the expasion

e^(it) = cos t+ i sin t

same can be used for other cube root

Answer 2

there is not any many times used thank you to discover the basis of a 6th degree polynomial. in fact, that's shown that there are such polynomials whose roots can't be expressed making use of the algebraic operations (+, -, *, /, powers and roots). many times used strategies in user-friendly terms exist as much as fourth degree. It enables that we are able to discover one root particularly easily: x = one million. Dividing the polynomial with the aid of (x - one million) supplies us x^5 + x^4 + x^3 + x^2 + x + 8 = 0 it extremely is a lot greater durable. I word, in the beginning, that different suggestions can't be constructive. because of the fact there are 5 roots and complicated roots are available in pairs, there could desire to be a minimum of one adverse actual root. If this root is -a, then -a^5 + a^4 - a^3 + a^2 - a + 8 = 0 or a^5 + a^3 + a = a^4 + a^2 + 8 If a will advance, the left facet will advance speedier than the superb facet. that's user-friendly to work out that a lies between one million and a pair of. that's no rational type, because of the fact the unique equation is monic and with out fractions. an actual answer isn't trouble-free to discover...

Answer 3

There are 6 distinct roots of this equation. Let y=x^3, to get y^2+7y-8=0. This factors as (y+8)(y-1)=0, so y=-8 or 1. Thus, x^3=-8 or x^3=1. Rewrite these:
x^3+8=0 or x^3-1=0.
Factor:
(x+2)(x^2-2x+4)=0 or (x-1)(x^2 +x+1)=0.
the first gives
x=-2, (2+- sqrt(-12))/2 and the second gives
x=1, (-1+-sqrt(-3))/2.

-2 and 1 are not repeated roots!!!!!!!

Answer 4

Your best bet is to let a variable y=x^3. Substitute so you get:
y^2 + 7y - 8 =0. The problem becomes much less intimidating.
(y + 8) (y - 1) = 0
y = -8 or y = 1
Substitute back, x^3 = -8 or x^3 = 1
x = -2 or x = 1

Answer 5

You can factor this equation.
(x^3 + 8)(x^3 -1) = 0
(x+2)(x^2 - 2x +4)(x-1)(x^2 +x +1) = 0
one of these x's must be zero to satisfy the equation
solve for x in x^2-2x+4 and x^2+x+1. Those will satisfy the initial equation. -2 and 1 will also satisfy it.

Answer 6

Boy did you post the wrong equation!

This one is quadratic in x^3. Make the substitution of z = x^3 and it is quadratic in z. Us the quadratic equation and:

z1 =( -7 + squareroot(49 - 4*(-8)))/2 = (-7 + squareroot(81))/2
z1 = 1
z2 = (-7 - squareroot(81))/2 = -8

The roots x are the cube roots of the zs:

x1 = 1
x2 = -2

Factoring out thesae two roots leaves: x^4 - x^3 + 2x + 4

Which can be sovled closed form as a quartic or factored into two quadratics: (x^2 + x + 1)(x^2 - 2x + 4)

Each quadratic has two roots giving:

x3 = (squareroot(-3) -1)/2
x4 = (-squareroot(-3) -1)/2
x5 = 1 + squareroot(-3)
x6 = 1 - squareroot(-3)

Answer 7

Very simple
Put
x^3=y
new equation is
y^2+7y-8=0
so y= -8 or 1
so x^3= -8 or x^3 = 1
or x = -2 or x=1
so the 6 roots are
x= -2, -2, -2, 1,1,1
satisfy now

Answer 8

first let x^3=z

therefore, z^2+7z-8=0
therefore, (z+8)(z-1)=0
z=-8 or z=1
x^3=-8 or x^3=1

x=-2 or x=1

The roots are repeated three times at -2 and 1 respectively, giving the 6 roots.

Answer 9

or the equation x^6 + 7x^3 - 8 = 0
WE can put the equation in the formula
(x³ +8 ) (x³ - 1) = 0
so we have
x³ = -8
and this has the roots
-2 , -2ω , -2ω²

AND
x³ = 1
and this has the roots
1 , ω , ω²

ω = (-1 + √3 i )/ 2
ω² = (-1 - √3 i )/ 2
i = √-1

Answer 10

y=x^3
y^2+7y-8=0
(y+8)(y-1)=0

(x^3+8)(x^3-1)=0
so x^3=-8, x=cube root of -8=-2
and x^3=1, so x=1.

Estimada Ana,
pues resulta que lo que me acabas de escribir es lo mismo que yo te escribi.
Tal vez lo que quieres decir, es que
x^3=-8 tiene 3 raices, una real, que es -2, y dos complejas que son:
2(cos(PI/3)+isin(PI/3))

y

2 ( cos (5PI/6)+isin(5PI/6))

y lo mismo con x^3=1,
la raiz real es x=1,
y las complejas son:

cos( 2PI/3)+isin(2PI/3)
cos(4PI/3 )+isin(4PI/3)

la manera de encontrarlas es bien sencilla.
1o. te fijas en cual es el valor absoluto de la raiz real
2o. divides 360 entre 3 (el orden de la raiz que quieres encontrar).
3o. a partir de la raiz que tienes mides el angulo que obtuviste arriba, tantas veces como sea necesario, en este caso 2 veces.
Utilizando el angulo las coordenadas estan dadas por el coseno y el seno.