Find the roots of the given cubic equation

[chwala]

Homework Statement

see attached

Relevant Equations

Ring Theory

Going through this, am still checking but will post all the same; which method did they apply to find the roots of the attachment below.

My thinking;

Let
$p+qi$

be the cube root of

$x^3-6x+2=0$

then,

$\sqrt{x(x^2-6)}=i\sqrt{2}$

$(p^2-q^2+2pqi)(p+qi)= x^3-6x+2$

We know that,

$\sqrt{x^3-6x}=0+ i\sqrt{2}$

...
$(p^3-pq^2)=0$

$\sqrt{(-3p^2q-2pq^2+q^3)}i =i\sqrt{2}$

 

[topsquark]

Homework Statement: see attached
Relevant Equations: Ring Theory

Going through this, am still checking but will post all the same; which method did they apply to find the roots of the attachment below.
View attachment 332895My thinking;

Let
$p+qi$

be the cube root of

$x^3-6x+2=0$

then,

$x(x^2-6)=i\sqrt{2}$

$(p^2-q^2+2pqi)(p+qi)= x^3-6x+2$

We know that,

$x^3-6x=0+ i\sqrt{2}$

...
$(p^3-pq^2)=0$

$(-3p^2q-2pq^2+q^3)i =i\sqrt{2}$

Where did you get
$x(x^2 - 6) = i \sqrt{2}$
from?

You are pretty much stuck with Cardano's method. I probably shouldn't give you this particular link, as it deals with your exact problem, but Cardano's method can be a bit tough to learn. (And this is an easier way to learn it than I used!) The idea is to write $x^3 - 6x + 2 = (a + b)^3$ in a useful way. See here.

-Dan

 

[chwala]

Where did you get
$x(x^2 - 6) = i \sqrt{2}$
from?

You are pretty much stuck with Cardano's method. I probably shouldn't give you this particular link, as it deals with your exact problem, but Cardano's method can be a bit tough to learn. (And this is an easier way to learn it than I used!) The idea is to write $x^3 - 6x + 2 = (a + b)^3$ in a useful way. See here.

-Dan

sorry i meant square root... i just amended. Let me look at your suggestion.

 

[martinbn]

The text in the picture does not suggest that it would be easy to find the roots. It only gives you the answer. Most likely it will be something they cover later on. What follows after that paragraph?

 

[anuttarasammyak]

Say the solutions are a,b,c
$$a+b+c=0$$
$$ab+bc+ca=6$$
$$abc=-2$$
We easily see that the given solution satisfies these conditions. It is not deriving but confirming the solution.

 

[WWGD]

Homework Statement: see attached
Relevant Equations: Ring Theory

Going through this, am still checking but will post all the same; which method did they apply to find the roots of the attachment below.
View attachment 332895My thinking;

Let
$p+qi$

be the cube root of

$x^3-6x+2=0$

then,

$\sqrt{x(x^2-6)}=i\sqrt{2}$

$(p^2-q^2+2pqi)(p+qi)= x^3-6x+2$

We know that,

$\sqrt{x^3-6x}=0+ i\sqrt{2}$

...
$(p^3-pq^2)=0$

$\sqrt{(-3p^2q-2pq^2+q^3)}i =i\sqrt{2}$

Sorry to nitpick, but didn't you mean $p+qi$ is _ a_ root of $x^3-6x+2$?

 

[chwala]

Sorry to nitpick, but didn't you mean $p+qi$ is _ a_ root of $x^3-6x+2$?

@WWGD I wanted it to be the cube root...I was thinking along that line...

 

[WWGD]

@WWGD I wanted it to be the cube root...I was thinking along that line...

Fair enough, but not sure what the cube root of an equation means. Do you mean $(p+qi)^{1/3}$ is a root of $x^3+ 3x+2$?

 

[chwala]

Fair enough, but not sure what the cube root of an equation means. Do you mean $(p+qi)^{1/3}$ is a root of $x^3+ 3x+2$?

That was my thinking based on the $3$ complex roots that are indicated...

 

[mathwonk]

Cardano is well explained by Euler. The upshot is to write your equation as
X^3 = 6X -2, and find u,v with 3uv = 6, i.e. uv =2, and u^3 + v^3 = -2. Then X = u+v.

Thus with A = u^3 and B = v^3, want A+B = -2 and AB = 8, which means Y=A and Y=B are solutions of Y^2 + 2Y +8 = 0, namely Y = -1 ± sqrt(-7).

Then u = cubrt(Y), and v = 2/u, give the 3 solutions.

The derivation is by assuming X = u+v, and expanding (u+v)^3 = 3uv(u+v) + u^3+v^3, setting u+v = X, 3uv = 6, u^3+v^3 = -2. Then X^3 = 6X -2 exactly when X = u+v, where u,v satisfy 3uv = 6, u^3+v^3 = -2.

 

[chwala]

Where did you get
$x(x^2 - 6) = i \sqrt{2}$
from?

You are pretty much stuck with Cardano's method. I probably shouldn't give you this particular link, as it deals with your exact problem, but Cardano's method can be a bit tough to learn. (And this is an easier way to learn it than I used!) The idea is to write $x^3 - 6x + 2 = (a + b)^3$ in a useful way. See here.

-Dan

I just had a look at the video; the Cardano's method is not difficult or tough to learn! its pretty straightforward. Just a question ...does it apply to all cubic functions of the form $ax^3+bx+c=0$? This approach is new to me... thanks though...

How do we manipulate her solution (shown below) to the three form of solutions indicated in post $1$?

$x=-\dfrac{2}{\sqrt [3]{1±i\sqrt{7}}}-{\sqrt [3]{1±i\sqrt{7}}}$

 

[mathwonk]

I tried to explain cardano's method briefly but completely above. I.e. as Euler explains in his elementary algebra book, since (u+v)^3 = 3uv(u+v) + (u^3+v^3),
is true for every u and v, we can always solve X^3 = AX + B, provided we can find u and v with 3uv = A and u^3+v^3 = B. I.e. then we need only set X = u+v.

But we know we can always solve for two numbers with given sum p and product q, just by solving the quadratic Y^2 - pY + q = 0. (Since in a quadratic, with roots r and s, we have (Y-r)(Y-s) = Y^2 - (r+s)Y + rs = 0, so the constant coefficient is the product rs, and the linear coefficient Is (minus) the sum r+s.)

In our case the two numbers we want to solve for are u^3 and v^3, since we know their sum and product, i.e. since 3uv = A, we know (uv)^3 = u^3v^3 = A/27, and we are given u^3 + v^3 = B.

So the first step in finding u and v is to find u^3 and v^3 by solving Y^2 - BY + A/27 = 0. Taking one such solution for Y to be u^3, we take all three cube roots to find all three values of u, and then to find v we use the fact that 3uv = A, so v = A/3u.
Then finally X = any of the three values of u+v.

Thus to solve X^3 = 6X -2, just set 3uv = 6, or uv = 2, so u^3.v^3 = 8, and u^3+v^3 = -2, so then solve Y^2 +2Y + 8 =0, for u^3 = Y = -1 - sqrt(-7), (you only need one solution since the other one is v^3.)

Then the three solutions for u are the three cube roots of this solution, and given any one of them, the others are found by multiplying by the two non trivial cube roots of 1.

In each case the corresponding values of v are v = 2/u. so one solution is
X = u + 2/u = cubert(-1 - sqrt(-7)) + 2/cubert(-1 -sqrt(-7)), and if t is a non trivial cube root of 1, i.e. since t^3-1 = (t-1)(t^2+t+1) = 0, if t is a solution of t^2+t+1 = 0, the other two solutions for X are:
X = tu + 2/tu, and X = t^2u + 2/t^2u.

To compare to the solutions in post #1, note that the quadratic formula gives
t = (1/2)(-1 ± i.sqrt(3)).

do you see it now?

 

[bamboum]

This equation has 3 real solutions . You can find it numerically -2.601 0.339 2.261. Are the answers of the text real values? Not evident.....

 

[chwala]

This equation has 3 real solutions . You can find it numerically -2.601 0.339 2.261. Are the answers of the text real values? Not evident.....

not sure of this...