Solution

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 \(x = -1.3\).