# Prove $|q|>64$: Roots of $x^{12}-pqx+p^2=0$

• MHB
• anemone
In summary, the absolute value of q must be greater than 64 in order to have real solutions for the equation $x^{12}-pqx+p^2=0$. This can be solved using the quadratic formula and the roots represent the possible values of q that satisfy the inequality $|q|>64$. There can be an infinite number of solutions for q as long as they satisfy the inequality.
anemone
Gold Member
MHB
POTW Director
Let $p$ and $q$ be real parameters. One of the roots of the equation $x^{12}-pqx+p^2=0$ is greater than 2. Prove that $|q|>64$.

Write the equation as a quadratic in $p$: $p^2 - qxp + x^{12} = 0$. For this to have a real solution for $p$, its discriminant $(qx)^2 - 4x^{12}$ must be nonnegative. So $q^2x^2 \geqslant 4x^{12}$. If $x>2$ then $x\ne0$ so we can divide by $x^2$ to get $q^2\geqslant 4x^{10} >4*2^{10} = 2^{12}$. Now take square roots to get $|q| > 2^6 = 64$.

## 1. What is the significance of the inequality $|q|>64$ in the equation $x^{12}-pqx+p^2=0$?

The inequality $|q|>64$ indicates that the roots of the equation must be greater than 64 in magnitude. This means that the solutions to the equation must be either positive or negative numbers that are greater than 64 or less than -64.

## 2. How does the value of $p$ affect the roots of the equation $x^{12}-pqx+p^2=0$?

The value of $p$ affects the roots of the equation in two ways. First, it determines the value of $q$ through the equation $pq=p^2$, which in turn affects the inequality $|q|>64$. Second, the value of $p$ also affects the coefficients of the equation, which can impact the nature of the roots (real or complex).

## 3. Can the equation $x^{12}-pqx+p^2=0$ have more than 12 distinct roots?

No, the equation can only have a maximum of 12 distinct roots. This is because the equation is of degree 12, meaning it can have at most 12 solutions. However, some of these solutions may be repeated or complex.

## 4. How can the inequality $|q|>64$ be proven for the equation $x^{12}-pqx+p^2=0$?

The inequality $|q|>64$ can be proven by using the discriminant of the equation, which is $D=p^2-4p^2=-3p^2$. Since the discriminant is negative, the equation only has complex roots. However, since the inequality requires real roots, we can conclude that $|q|>64$ must be true.

## 5. Can the inequality $|q|>64$ be satisfied if $p$ is a negative number?

Yes, the inequality $|q|>64$ can be satisfied if $p$ is a negative number. This is because the inequality only requires the magnitude of $q$ to be greater than 64, not the actual value. Therefore, as long as $p$ and $q$ satisfy the equation $pq=p^2$ and the resulting value of $q$ has a magnitude greater than 64, the inequality will be satisfied.

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