Advent calendar 2021
19 December
The equation \(352x^3-528x^2+90=0\) has three distinct real-valued solutions.
Today's number is the number of integers \(a\) such that the equation
\(352x^3-528x^2+a=0\) has three distinct real-valued solutions.
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The function \(f(x)=352x^3-528x^2+a\) with \(a=90\) looks like this (click to enlarge):
Adjusting the value of \(a\) will move the curve up and down. The equation \(f(x)=0\) will have three distinct solutions as long as the local minimum is below the \(x\)-axis
and the local maximum is above the \(x\)-axis.
The minimum and maximum are at \(x=1\) and \(x=0\). The difference between \(f(1)\) and \(f(0)\) is 176. This means that there are 175 possible integer values of \(a\) (as the endpoints are not included).
![](javascript:showlimage("advent2021-cubic.png"))