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Given: $P(x)=x^4 -4(m+2)x^2 + m^2$ has 4 real roots.

Show: the sum of the possible least 3 integer values of $m$ is zero.

This is a question asked in an entrance exam (Colégio Naval 92).

My attempt: Let $x^2=y$ to get $Q(y)=y^2 -4(m+2)y + m^2=$ and find conditions on $m$ for real roots for $Q(y)$. The discriminant would be defined by $$\vartriangle=16(m+2)^2-4m^2=4\left(4(m^2+4m+4)-m^2\right)=4(3m^2+16m+16).$$ Therefore, for real roots, we need $\vartriangle\ge 0$ or $$3m^2+16m+16\ge 0\Leftrightarrow 9m^2+48m+64-64+48\ge 0\Leftrightarrow(3m+8)^2\ge 16$$ leading to the conditions for $m$: (a) $3m+8\ge 4$ or (b) $3m+8\le -4$, or developing both expressions, (a) $m\ge -4/3$ or (b) $m\le -4$, for $\vartriangle\ge 0$.

But that is not enough. Four real roots for $P(x)$ requires that all roots for $Q(y)$ are not only real, but positive. Therefore, we need, in addition, $$\frac{1}{2}\left(4(m+2)\pm 2\sqrt{3m^2+16m+16}\right)=2(m+2)\pm \sqrt{3m^2+16m+16}\ge 0$$

My difficulty is on how to solve this last condition for positive roots for $Q(y)$. Once I have the set of values for $m$ from this condition, next step would be solving for the intersection of this set with the previous condition for real roots, $m\ge -4/3$ or $m\le -4$. With the resulting set I would check for the sum of the least 3 integers to show that it is zero.

Hints or full answers are welcome. I really don't know if my approach is most appropriate way to solve this problem.

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EDIT: I realized that calling Macavity's answer "working by accident" is not fair. What it does is use the general prescription for solving an inequality involving one irrational expression, just doesn't explicitly state that. In general, $f(x) \geq \sqrt{g(x)}$ is equivalent to $$f(x) \geq 0 \;\; \land \;\; (f(x))^2 \geq g(x) \geq 0 $$ and in your case only the first of the above yields a new restriction on $m$. Be careful though, if the inequality is the other way, it is more complicated. $f(x) \leq \sqrt{g(x)}$ is equivalent to $$ \{ f(x) < 0 \;\;\land\;\; g(x) \geq 0 \} \;\;\lor\;\; \{ f(x) \geq 0 \;\;\land\;\; (f(x))^2 \leq g(x) \} $$


There is another very powerful approach for imposing conditions on the roots of a quadratic by considering its graph. Let $Q(x) = ax^2 + bx + c$. It has two non-negative $x$-intercepts if and only if $$ \begin{align} \Delta \geq 0 \\ aQ(0) = ac \geq 0 \\ \frac{-b}{2a} \geq 0 \end{align} $$ The first is equivalent to having real roots (if you want distinct, make it strict). The second says that you want $0$ to be outside of the interval between the roots (which happens iff $Q(0) \geq 0$ for positive $a$ and iff $Q(0) \leq 0$ for negative $a$). The last one makes sure that if it is outside, it is on the left side, because it will be to the left of the vertex. So together they are equivalent to saying that $0$ is less than or equal to both roots (if you want strictly positive roots, make the second inequality strict).

It is easily adapted to when other configurations are desired:

  • for two negative roots, flip the direction of the last one;

  • for one positive and one negative, flip the second one and get rid of the third one altogether (in fact, the first one also becomes redundant in this case);

  • if you want roots bigger/smaller than some particular number $t$ instead of $0$, replace $Q(0)$ with $Q(t)$ in the second and $0$ with $t$ in the third.

  • it is even possible to apply the same method (carefully) to situations where two (or even more) numbers are involved in the conditions on the roots, such as if we want both roots in some interval $(s, t)$, or one inside and one outside (possibly on a particular side), or each in its own interval, etc.

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  • $\begingroup$ Nick, great insight (+1). Very helpful indeed and easy to remember. $\endgroup$ – bluemaster Oct 22 '17 at 19:03
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May be, you could consider the function $$P(x)=x^4 -4(m+2)x^2 + m^2$$ $$P'(x)=4x^3 -8(m+2)x$$ $$P''(x)=12x^2 -8(m+2)$$

The first derivative cancels at $$x_1=-\sqrt{4+2m} \qquad x_2=0\qquad x_3=\sqrt{4+2m}$$ For these points $$P(x_1)=-3 m^2-16 m-16 \qquad P(x_2)=m^2 \qquad P(x_3)=-3 m^2-16 m-16$$

So, in order to have four distinct roots, from the values of $P(x_i)$ we need to have $m \neq 0$ and $3m^2+16m+16>0$. The last condition excludes the range $-4 \leq m \leq -\frac 43$.

Since $x_1$ and $x_3$ must be minimum points and $x_2$ a maximum point, we also need $m>-2$ since $$P''(x_1)=16(m+2) \qquad P''(x_2)=-8(m+2) \qquad P''(x_3)=16(m+2)$$ This seems to define quite well the range of acceptable values of $m$.

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  • $\begingroup$ @Macavity. Correct ! it is $m \neq 0$. Thanks. $\endgroup$ – Claude Leibovici Oct 22 '17 at 4:39
  • $\begingroup$ Even when $m=0$, we have a double root at $x=0$ and two other distinct real roots. Arguably this is still four real roots, counting multiplicity, though not specified clearly by OP. +1 anyway. $\endgroup$ – Macavity Oct 22 '17 at 4:44
  • $\begingroup$ @Macavity. I totally agree with you. We can treat the cases of multiple roots, each one adding conditions. $\endgroup$ – Claude Leibovici Oct 22 '17 at 4:52
  • $\begingroup$ Claude, thanks for the nice development (+1). Do you guys (Claude and @Macavity) believe there is a way to solve without using calculus arguments? This question appeared in an entrance exam that assumes knowledge up to the 9th grade. I think that is a hard question at this level. $\endgroup$ – bluemaster Oct 22 '17 at 10:02
  • $\begingroup$ @bluemaster have added an answer completing your approach. $\endgroup$ – Macavity Oct 22 '17 at 11:16
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For a calculus less approach, we can complete your analysis. As you noted, the roots of $Q(y)$ are real if $$3m^2+16m + 16 \geqslant 0 \implies m \in (-\infty, -4] \cup [-\tfrac43, \infty)$$

Then for them to be non-negative, we just need $m+2 \geqslant 0 \iff m \geqslant -2$ as then $$2(m+2) \geqslant \sqrt{3m^2+16m+16} \iff (2(m+2))^2 \geqslant 3m^2+16m+16 \iff m^2 \geqslant 0$$

Thus all the roots are real and non-negative iff $m \geqslant -\tfrac43$, the lowest three integers possible then are $m=-1, 0, 1$...

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