Show that the equation \(x^3 - 6x^2 + 12 = 0\) has two positive roots and one negative root.
Let \(f(x) = x^3 - 6x^2 + 12\), which is a cubic, and thus has at most three real roots. We can tabulate various values for \(f(x)\).
| \(x\) | \(f(x)\) |
|---|---|
| -2 | -20 |
| -1 | 5 |
| 0 | 12 |
| 1 | 7 |
| 2 | -4 |
| 3 | -15 |
| 4 | -20 |
| 5 | -13 |
| 6 | 12 |
The function \(f(x)\) changes sign three times in the table, so that \(f(x)\) has three real roots, one between \(x = -2\) and \(x = -1\); one between \(x = 1\) and \(x = 2\); and a third between \(x = 5\) and \(x = 6\).
So \(f(x)\) has two positive roots and one negative root.
Show that the negative root lies between \(-1\) and \(-1.5\)…
We saw above that \(f(-1) = 5\). As \(f(-1.5) = -\dfrac{39}{8} < 0\), the negative root must lie between \(x = -1\) and \(x = -1.5\).
…and find it correct to one decimal place.
We can tabulate more values for \(f(x)\).
| \(x\) | \(f(x)\) |
|---|---|
| -1.5 | -4.875 |
| -1.4 | -2.504 |
| -1.3 | -0.337 |
| -1.2 | 1.632 |
| -1.1 | 3.409 |
| -1 | 5.0 |
Thus the negative root lies between \(x = -1.3\) and \(x = -1.2\).
We have \(f(-1.25) = 0.671875 > 0\), and so the root falls between \(x = -1.3\) and \(x = -1.25\), so that, to one decimal place, the negative root is \(x = -1.3\).
