The equation \(x^3 + 10x^2 – 100x + 1729 \) has at least one complex root α such that |α| > 12.

**False ****

**Discussion: **

*A fun fact : 1729 is the Ramanujan Number; it is the smallest number expressible as the sum of two cubes in two different ways*

We conduct normal extrema tests. First derivative is \(3x^2 + 20x – 100 \). Critical points are (zeroes of first derivative) are -10 and 20/6. Second Derivative (6x + 20) test tells that -10 is the (x coordinate of ) maxima and 20/6 is minima. Since these maxima and minima turns out to be global extremas, we check f(20/6) and find that it is positive. Hence the function has exactly one real root smaller than -10 (other wise the global minima would have been negative).

Product of the roots is – 1729. Since the coefficients are all real we conclude the complex roots must have occurred in conjugates. Let \(\alpha \) be a complex root and t be the real root. Then \(|\alpha|^2 t = -1729 \) . We wish to show that \(| \alpha | > 12 \implies | \alpha |^2 > 144 \implies | \alpha |^2 t < 144t \) since t is negative. This says that -13 < t but plugging in -13 in f(x) we see that it is positive. Since the function is monotonically increasing up to -10, the real root must have occurred before -13 (graphing shows that it occurs somewhere between -20 and -19). Hence the claim is false.

*Related*

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