Relationships between the roots

[Bruno Tolentino]

A quadratic equation in this format [tex] x² - 2 A x + B² = 0 [/tex]

where:

A = (a + b)×(1/2)
B = (a × b)^(1/2)

a and b are the roots and the roots can be expressed in terms of midpoint and radius

u = midpoint
v = radius

a = u + v
b = u - v

can be modified and expressed like: x² - 2 (u) x + (u² - v²) = 0.

The roots x1 and x2 are therefore: [tex]x_1 = x_1(u,v) = u + v[/tex] [tex]x_2 = x_2(u,v) = u - v[/tex] Or: [tex]x_1 = x_1(a, b) = \frac{a+b}{2} + \frac{|a-b|}{2}[/tex] [tex]x_2 = x_2(a, b) = \frac{a+b}{2} - \frac{|a-b|}{2}[/tex]

Until here, these relationships are known. No problem! The problem begins when I ask: "how are these relationships for the cubic equation?"

I know that analog cubic equation is [tex] x^3 - 3 A x^2 + 3 B^2 x - C^3 = 0 [/tex] and that
[tex] A = \frac{a+b+c}{3} [/tex] [tex] B = \sqrt[2]{\frac{ca + ab + bc}{3}} [/tex] [tex] C = \sqrt[3]{abc} [/tex]
a, b and c are the roots. The roots can be called too of x1, x2 and x3. But, by analogy with the relationships above, how I can to express x1, x2 and x3 in terms of a linear combination of a, b and c?
[tex]x_1 = x_1(a, b, c) = ? [/tex] [tex]x_2 = x_2(a, b, c) = ? [/tex] [tex]x_3 = x_3(a, b, c) = ? [/tex]

And how express x1, x2 and x3 in terms of u, v and w?

[tex]x_1 = x_1(u, v, w) = ? [/tex] [tex]x_2 = x_2(u, v, w) = ? [/tex] [tex]x_3 = x_3(u, v, w) = ? [/tex]

 

[pasmith]

See Lagrange's Method.

 

[mathwonk]

I usually look at these this way:

QUADRATIC FORMULA:
To solve a quadratic equation X^2 -bX + c = 0, first assume the solutions are X=r and X=s, and note that then (by the root-factor theorem), the equation factors as X^2 -bX + c = (X-r)(X-s) = X^2 -(r+s)X + rs, so that we must have b = r+s and c = rs. Thus we know in particular the sum b of the roots. Hence if we only knew the difference, say d = r-s, of the roots, we would be done, since then we would have 2r = b+d and 2s = b-d.

So we assume the roots are expressed as a sum, namely r = b/2 + d/2, and s = b/2 - d/2. Then c = rs = b^2/4 - d^2/4, so d^2 = b^2-4c.

Thus 2r = b + sqrt(b^2-4c) and 2s = b - sqrt(b^2-4c), the usual formula, if you notice we had a minus sign in the linear term of our original equation.CUBIC FORMULA:
To solve a cubic equation, we start with a simplified one of form X^3 -3bX + c = 0, and again assume we want to find X as a sum X = (p+q). Plugging in gives (p+q)^3 = 3b(p+q) + c, and expanding gives p^3 + 3p^2q + 3pq^2 + q^3 = p^3 + q^3 + 3pq(p+q) =
3b(p+q) + c, and for this to hold means that pq = b, and c = p^3+q^3. Cubing the first of these gives p^3q^3 = b^3, and p^3+q^3 = c. Since we know b and c, we know both the sum and the product of the cubes p^3 and q^3. Can we find p^3 and q^3 from this? If so, then we could take cube roots and find p and q, and finally add them and get our root X = p+q.

Just recall in a quadratic equation of form X^2 - BX + C, that B and C are precisely the sum and product of the desired roots, and we can find those roots from B and C. I.e. we can find any two numbers when we know their sum and product, by solving a quadratic.

Since p^3+q^3 = c and p^3q^3 = b^3, the numbers p^3 and q^3, which can be used to give a solution X = p+q of our cubic, are solutions of the quadratic equation
t^2 -ct + b^3 = 0e.g. to solve X^3 = 9X + 28, we have b = 3, c = 28, and so we solve t^2 -28t + 27 = 0. Here B^2-4C = 676, whose square root is 26, so we get t = (1/2)( 28 ± 26) = {27, 1}, for p^3 and q^3, so p,q are 1 and 3, and hence X=1+3 = 4 solves the cubic. Of course if we know about complex numbers, there are two more cube roots of 1 and 27, and we get two more complex roots. (Only two more because b = pq, so we must always have q = b/p, i.e. the choice of the cube root q is determined by the choice of p.)

Finally, one can translate the variable in any cubic equation to change it into one with zero quadratic term, so this process works in general.

 

[Bruno Tolentino]

See Lagrange's Method.

See my question in this topic: https://www.physicsforums.com/threads/3x3-matrix.851891/

My question is the same and I still I don't know the answer...