Third Degree Burn

Calculus Level pending

If a third degree polynomial with real coefficients has two distinct real roots $a$ and $b$ , then which of the following is true?

Note: Consider a complex root to be a number $x$ such that $x = w + yi$ , where $y \neq 0$ and $i = \sqrt{-1}$ .

f \left ( \frac{a+b}{2} \right )> 0

f'(a)

can never be a perfect integer square The polynomial has one complex root

f'(a) \cdot f'(b) = 0

1 solution

Guilherme Dela Corte
Jan 28, 2017

Relevant wiki: Derivatives of Polynomials - Basic

$f$ is of the form $f(x) = k(x-a)^2(x-b)$ or $f(x) = k(x-a)(x-b)^2$ , where $k,a,b$ are real numbers. Because $f$ has real coefficients and already has all of its roots defined ( $a,a,b$ or $a,b,b$ ), it cannot yield $f(x_0) = 0$ such that $x_0$ is "complex".

Picking $f(x) = k(x-a)(x-b)^2$ , we note that $f'(x) = 2k(x-a)(x-b) + k(x-b)^2 \Rightarrow f'(a) = k(a-b)^2, \; f'(b) = 0$ . Let $k = 1, \; a = 2, \; b = 3$ . We see that $f'(2) = 1$ , which is a perfect integer square. It is easy to see that $f'(a) \cdot f'(b) = 0$ .

Picking $k = 1, a = 3, b = 1$ , we see that $f(2) = -1.125 < 0$ .

Third Degree Burn

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