Find $m$ if $x^3 - mx^2 - 4 = 0$ and $x^3 + mx + 2 =0$ have a common root.

Question

Good Day

Today, I was trying to solve this problem below:

Find the values real number $m$ can take if $x^3 - mx^2 - 4 = 0$ and $x^3 + mx + 2 =0$ have a common root.

Here's what I did:

Assume the common root to be $a$. Substituting:

$$(1)\ a^3 - ma^2 - 4 = 0$$

$$(2)\ a^3 + ma + 2 = 0$$

Subtracting $(1)$ from $(2)$ gives:

$$ma^2 + ma + 6 = 0$$

Using this as a quadratic in $a$, we use the fact that $D \geq 0$ because $a$ is real.

Thus, $$-4(m)(6) + (m)^2 \geq 0$$

So, $$m(m - 24) \geq 0$$

which is true precisely when $m \in (-\infty, 0]$ $\cup$ $[24, \infty)$

However, the answer given to me is only $m = -3$. Where am I going wrong?

Thanks!

Answer 1

Assume the common root to be $a$.

Hint: $\;a \ne 0\,$, then $\,m=\dfrac{a^3-4}{a^2}=- \dfrac{a^3+2}{a}\,$. The latter equality gives $\,a^4+a^3+2a-4=0\,$ (n.b. edited to fix typo).

Answer 2

The resultant of two univariate polynomials is zero iff they have a common root. In this case, the two polynomials' resultants with respect to $x$ is $$4m^4+8m^3-36m^2+216=4(m+3)^2(m^2-4m+6)$$ The quadratic in $m$ has no real roots, so $m=-3$ is the only solution.

Answer 3

From $(2)$ we get

$$(3)\ a^4 + ma^2 + 2a = 0.$$

Adding $(1)$ and $(3)$ gives

$$(4) \ a^4+a^3+2a-4=0.$$

It is easy to see that the real solutions of $(4)$ are $1$ and $-2$. Now show that $(1)$ and $(2)$ give a solution $m$ only if $a=-2.$ This gives $m=-3.$

Answer 4

When you subtracted (1) from (2), you turned two constraints into one. You need to get another constraint from a different combination of (1) and (2).

$2\times(2) + (1)$ is a good choice as it would also result in a quadratic (since $a\ne0$):

$$3a^2 - ma + 2m = 0$$ $$ma^2 + ma + 6 = 0$$

Now add the two together to get $(m+3)(x^2+2)=0$, from which it follows that $m=-3$.

Answer 5

The intersections equation you found, $ \ mx^2 + mx + 6 \ = \ 0 \ , $ is useful, but does not by itself tell us about the zeroes of the polynomial; rather, it only locates the $ \ x-$coordinates of the two intersection points between the curves for $ \ x^3 - mx^2 - 4 \ \ $ and $ \ x^3 + mx + 2 \ \ . $ These points lie at $$ x \ \ = \ \ \frac{-m \ \pm \ \sqrt{m^2 \ - \ 24m}}{2m} \ \ = \ \ -\frac12 \ \pm \ \frac{\sqrt{1 \ - \ \frac{24}{m}}}{2} \ \ . $$ As you found, this means that the two function curves intersect only for $ \ m \ \ge \ 24 \ \ $ or $ \ m \ < \ 0 \ \ . $ (We can discard $ \ m = 0 \ $ as this produces the two "parallel" cubic curves $ \ y \ = \ x^3 + 2 \ \ $ and $ \ y \ = \ x^3 - 4 \ \ . ) $

A differing approach we may take is to consider the properties of the functions and their curves and what this quadratic relation tell us. Something we can say immediately (although it is not clear how it is helpful) is that the common zero $ \ r \ $ must be a common factor of $ \ 4 \ $ and $ \ -2 \ \ . $ If we "complete-the-cube" for the function $ \ x^3 + bx^2 + cx + d \ , \ \ $ we obtain $$ \ \left(x + \frac{b}{3} \right)^3 \ + \ \left(c - \frac{b^2}{3} \right)x \ + \ \left( d - \frac{b^3}{27} \right) \ \ , $$ which indicates a symmetry axis (neglecting "vertical displacement") at $ \ x \ = \ -\frac{b}{3} \ $ and that the curve will have local extrema ("turning-points") if $ \ c < \frac{b^2}{3} \ \ . $ Consequently, $ \ y \ = \ x^3 - mx^2 - 4 \ $ has a symmetry axis about $ \ x \ = \ \frac{m}{3} \ \ $ and always has turning-points since $ \ c = 0 \ \ ; $ the curve for $ \ y \ = \ x^3 + mx + 2 \ $ has a symmetry about the $ \ x-$axis $ \ ( b = 0 ) \ $ and turning-points when $ \ m < 0 \ \ . $ [Calculus would enable us to say rather more, but we are limiting ourselves to algebra of polynomials and analytic geometry.] We will be able to narrow down possibilities with this information.

It might also be noted that the Rule of Signs indicates that:

• for $ \ m > 0 \ , \ \ x^3 - mx^2 - 4 \ $ has one positive real zero and no negative real zeroes; $ \ x^3 + mx + 2 \ $ has no positive real zeroes and one negative real zero

• for $ \ m < 0 \ , \ \ x^3 - mx^2 - 4 \ $ has one positive real zero and two or no negative real zeroes; $ \ x^3 + mx + 2 \ $ has two or no positive real zeroes and one negative real zero

So the Rule of Signs analysis permits us to reject the case for $ \ m \ \ge \ 24 \ \ . $ This is confirmed by investigating the function values. For $ \ m = 24 \ \ , $ the single intersection at $ \ x = -\frac12 \ $ gives us $ \ y \ = \ -\frac{81}{8} \ \ . $ As $ \ m \ $ increases, the intersection bifurcates into two points with $ \ x \rightarrow -1^{+} \ $ and $ \ x \rightarrow 0^{-} \ \ . $ For the former intersection, $ \ y \rightarrow -\infty \ \ , $ while the latter approaches the $ \ y-$intercept at $ \ ( 0 \ , -4 ) \ \ . $ Hence, no common zero is possible for this case.

In a similar fashion, we can show that $ \ m \ $ cannot have a large negative value if there is to be a common zero for the two polynomials. (The Rule of Signs leaves the possibility open for $ \ m < 0 \ \ . ) $ If we choose a convenient value of $ \ m = -8 \ \ , $ the intersections are found at $ \ x = -\frac32 \ $ and $ \ x = +\frac12 \ \ , $ with the corresponding function values $ \ y = \frac{85}{8} \ $ and $ \ -\frac{15}{8} \ \ ; $ we find that the S-turns in the two cubic curves are intertwining. As $ \ m \ $ decreases from this value, the $ \ x-$coordinates of the intersections behave as $ \ x \rightarrow -1^{-} \ $ and $ \ x \rightarrow 0^{+} \ \ , $ with the former intersection having $ \ y \rightarrow +\infty \ $ and the other intersection again approaching $ \ ( 0 \ , -4 ) \ \ . $ So there can be no common zero for this situation either.

Thus, if a common zero between the polynomials exists, we must have $ \ -8 < m < 0 \ \ . $ We could continue to analyze the behavior of the intersections, but we see that they must lie not far from the origin. So we could just try common factors of $ \ 4 \ $ and $ \ -2 \ $ at this point to see what happens.

It might be remarked here that there is something suspicious about these real zeroes. The Rule of Signs shows that there is one positive and one negative real zero among the two polynomials. But a cubic polynomial with real coefficients must have either one or three real zeroes: could it be that there is a double zero?

On testing possible integer zeroes, we observe for the intersections that

$ \mathbf{x = +1 \ \ } \Rightarrow \ \ 1 \ - \ \frac{24}{m} \ = \ 9 \ \ \Rightarrow \ \ m \ = \ -3 \ \ ; $

$ \mathbf{x = -1 \ \ } $ [not admissible: asymptotic value] ;

$ \mathbf{x = +2 \ \ } \Rightarrow \ \ 1 \ - \ \frac{24}{m} \ = \ 25 \ \ \Rightarrow \ \ m \ = \ -1 \ \ ; $

$ \mathbf{x = -2 \ \ } \Rightarrow \ \ 1 \ - \ \frac{24}{m} \ = \ 9 \ \ \Rightarrow \ \ m \ = \ -3 \ \ . $

For the $ \ m = -1 \ $ case, the function curves intersect at $ \ (2 \ , 8) \ \ , $ so $ \ x = 2 \ $ is not a zero of either function. On the other hand, for $ \ m = -3 \ $ , we find two intersections at $ \ x = -2 \ $ and $ \ x = +1 \ \ ; $ we find the function values to be

$ \mathbf{f(x) \ = \ x^3 \ + \ 3x^2 \ - \ 4 \ \ : } \quad f(-2) \ = \ -8 + 12 - 4 \ = \ 0 \ \ \ , \ \ \ f(1) \ = \ 1 + 3 - 4 \ = \ 0 \ \ ; $

$ \mathbf{g(x) \ = \ x^3 \ - \ 3x \ + \ 2 \ \ : } \quad g(-2) \ = \ -8 + 6 + 2 \ = \ 0 \ \ \ , \ \ \ f(1) \ = \ 1 - 3 + 2 \ = \ 0 \ \ . $

[More thorough analysis would show that the intersections move away from the $ \ x-$axis as $ \ m \ $ is "shifted away" from $ \ -3 \ \ . ] $

Polynomial or synthetic division shows that $$ x^3 \ + \ 3x^2 \ - \ 4 \ \ = \ \ (x - 1) · (x + 2)^2 \ \ \ \text{and} \ \ \ x^3 \ - \ 3x \ + \ 2 \ \ = \ \ (x - 1)^2 · (x + 2) \ \ , $$

which means that the singleton zero of each of these polynomials makes an intersection with the double zero of the other (an initially unanticipated property!). The only solution for the problem then is $ \ \mathbf{m \ = \ -3 \ \ } . $