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anemone
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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 $|q|>64$.
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.
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).
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.
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.
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.