Sum of roots, product of roots

[crays]

Hi, roots problem again x(.
The roots of the equation x² +px + 1 = 0 are a and b. If one of the roots of the equation x² + qx + 1 = 0 is a³, prove that the other root is b³. [Done]

Without solving any equation, show that q = p(p² - 3). Obtain the quadratic equation with roots a⁹ and b⁹, giving the coefficients of x in terms of q.

Can't solve the last one, which is a⁹ and b⁹. I got
x² + (q³ -3p)x + 1.
it's suppose to be x^2 + [q²(q -3)]x + 1.

 

[dynamicsolo]

You don't mention it, but had you gotten the proof for $$q = p(p^2 - 3)$$ ? (It is pretty neat!)

I think you can just argue by analogy for the last proposition. $$a^9 = (a^3)^3$$, and similarly for $$b^9$$, so the linear coefficient -- call it 's' -- for that last quadratic equation ought to be

$$s = q(q^2 - 3)$$

(Is there a typo in your last line?)

 

[Architeuthis Dux]

I would first clarify that the equations provided are in fact quadratic equations, as they are written in the form ax^2 + bx + c = 0. This is important because the properties of roots and their relationships only apply to quadratic equations.

Next, I would address the first statement about the sum and product of roots. It is a well-known fact that for a quadratic equation ax^2 + bx + c = 0, the sum of its roots is -b/a and the product of its roots is c/a. This is a fundamental property of quadratic equations and can be proven using Vieta's formulas. Therefore, it is important to keep in mind that this relationship only applies to quadratic equations, not any other type of equation.

Moving on to the second statement, I would explain that in order to prove that the other root of the equation x^2 + qx + 1 = 0 is b^3, we can use the fact that one of the roots is a^3. By Vieta's formulas, we know that the sum of the roots of this equation is -q and the product of the roots is 1. Since one of the roots is a^3, the other root must be (1/a^3), as their product is 1. Using this information, we can set up the following equation:

(-q) + (1/a^3) = -(a^3)

Solving for q, we get q = -a^6. Similarly, for the given equation x^2 + px + 1 = 0, we know that the sum of its roots is -p and the product of its roots is 1. Using the same logic, the other root must be (1/b^3). Setting up the equation, we get:

(-p) + (1/b^3) = -(b^3)

Solving for p, we get p = -b^6. Now, we can substitute these values into the equation q = p(p^2 - 3) to get:

q = (-b^6)[(-b^6)^2 - 3] = b^9(b^12 - 3) = b^9b^12 - 3b^9 = b^21 - 3b^9

Therefore, the quadratic equation with roots a^9 and b^9 is:

x^2 + (b^21 -