Solution of equations having multiple roots

Any root which occurs more than once in an equation is called a multiple root. Thus $3$, $3$, $3$, $-2$ are the roots of

$$x^4 - 7x^3 + 9x^2 + 27x - 54 = 0;$$

hence $3$ is a multiple root occurring three times. Evidently this equation may also be written in the form

$$(x - 3)^3(x + 2) = 0.$$

Let $f(x)$ denote an integral rational function of $x$ having a multiple root $a$, and suppose it occurs $m$ times. Then we may write

$$f(x) = (x-a)^m\phi (x),$$ (6.18)

where $\phi(x)$ is the product of the factors corresponding to all the roots of $f(x)$ differing from $a$. Differentiating (6.18),

$$f'(x) = (x-a)^m\phi '(x) + m\phi (x)(x - a)^{m - 1},$$

or,

$$f'(x) = (x-a)^{m - 1}[(x - a)\phi '(x) + m\phi (x)].$$ (6.19)

Therefore $f'(x)$ contains the factor $(x - a)$ repeated $m - 1$ times and no more; that is, the highest common factor (H.C.F.) of $f(x)$ and $f'(x)$ has $m - 1$ roots equal to $a$.

In case $f(x)$ has a second multiple root $\beta$ occurring $r$ times, it is evident that the H.C.F. would also contain the factor $(x -\beta)^{r - 1}$ and so on for any number of different multiple roots, each occurring once more in $f(x)$ than in the H.C.F.

We may then state a rule for finding the multiple roots of an equation $f(x) = 0$ as follows:

• FIRST STEP. Find $f'(x)$.

• SECOND STEP. Find the H.C.F. of $f(x)$ and $f'(x)$.

• THIRD STEP. Find the roots of the H.C.F. Each different root of the H.C.F. will occur once more in $f(x)$ than it does in the H.C.F.

If it turns out that the H.C.F. does not involve $x$, then $f(x)$ has no multiple roots and the above process is of no assistance in the solution of the equation, but it may be of interest to know that the equation has no equal, i.e. multiple, roots.

Example 6.10.1   Solve the equation $x^3 - 8x^2 + 13x - 6 = 0$.

Solution. Place $f(x) = x^3 - 8x^2 + 13x - 6$.

First step. $f'(x) = 3x^2 - 16x + 13$.

Second step. H.C.F. = $x-1$.

Third step. $x - 1 = 0$, therefore $x=1$.

Since 1 occurs once as a root in the H.C.F., it will occur twice in the given equation; that is, $(x - 1)^2$ will occur there as a factor. Dividing $x^3 - 8x^2 + 13x - 6$ by $(x - 1)^2$ gives the only remaining factor $(x - 6)$, yielding the root $6$. The roots of our equation are then $1$, $1$, $6$. Drawing the graph of the function, we see that at the double root $x=1$ the graph touches the $x$-axis but does not cross it.

Note: Since the first derivative vanishes for every multiple root, it follows that the $x$-axis is tangent to the graph at all points corresponding to multiple roots. If a multiple root occurs an even number of times, the graph will not cross the $x$-axis at such a point (see Figure 6.11); if it occurs an odd number of times, the graph will cross.

Figure 6.11: plot of $f(x) = (x-1)^2(x-6)$ illustrating a multiple root.