I am given the following equation:


And I have to find the values of the parameter $m$ for which the roots of the equation are all real.

The possible answers are:

A. $m=0$

B. $1 \le m \le 2$

C. $-1 \le m \le - \dfrac{1}{2}$

D. $m \in \emptyset$

E. $m > \dfrac{1}{2}$

I know that if it would've been a parabola (so instead of $x^4$, we'd have $x^2$ and instead of $x^2$ we'd have $x$) we could solve this by making sure that the minimum of the parabola $V(- \dfrac{b}{2a}, - \dfrac{\Delta}{4a})$ has its $y$ value, $- \dfrac{\Delta}{4a}$, always below the $Ox$ axis or on the $Ox$ axis (in which case we'd have a double real root). But the fact that we don't have a parabola put me in the dark. I thought about doing some substitution like $t=x^2$ and turn it into a parabola but then I don't know how to turn the solutions of the created parabola into the solutions of the initial equation.


Substitutions are the way to go. Let $t=x^2$ and we have the equation $$t^2+(2m-1)t+2m+2=0$$

In addition to making sure that we have two real roots for this equation, we also need to make sure that both of these roots are non-negative. As such, we need $$2m-1\leq 0\Rightarrow m\leq\frac12$$ and $$2m+2 \geq 0\Rightarrow m\geq-1.$$

Simplified, we have $m\in[-1,\frac12]$.

Calculating the discriminant to have real roots:

\begin{align} \Delta=b^2-4ac&\geq0\\ (2m-1)^2-4(1)(2m+2)&\geq0\\ 4m^2-4m+1-8m-8&\geq0\\ 4m^2-12m-7&\geq0\\ (2m+1)(2m-7)&\geq0 \end{align} And we have $\Delta\geq0$ when $m\in(-\infty,-\frac12]\cup[\frac72,\infty)$.

From the intervals that we have for $m$, the largest intersection we have is $m\in[-1,-\frac12]$, which is answer $\boxed{\text{C}}$.

  • $\begingroup$ I must be forgetting something - why do you need to make sure that both of the roots are nonnegative? $\endgroup$ – Axion004 Oct 21 '19 at 1:52
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    $\begingroup$ @Axion004 Still need to sub back in $t=x^2$. No real roots for $x$ if $t$ is negative. $\endgroup$ – Andrew Chin Oct 21 '19 at 1:55
  • $\begingroup$ Also, why does $2m−1\le 0$ for the roots to be nonnegative? $\endgroup$ – Axion004 Oct 21 '19 at 2:51
  • $\begingroup$ @Axion004 in $a(x-\alpha)(x-\beta)=ax^2+bx+c=0$, we have negative $\alpha+\beta=b$ and positive $\alpha\beta=c$ for $\alpha,\beta>0$. $\endgroup$ – Andrew Chin Oct 21 '19 at 2:58
  • $\begingroup$ @AndrewChin Let me know if I got this correctly. We do the substitution $t=x^2$ and then find the values of $m$ for which this new equation, with $t$'s, has all of its roots real. The roots of the initial equation are the square roots of the $t$ roots. So, for the original roots to be real, we also need to make sure that the $t$ roots are non-negative. To make sure the $t$ roots are non-negative we make create the conditions: $S \ge 0$ and $P \ge 0$, where we have $S = t_1 + t_2 = - \dfrac{b}{a}$ and $P = t_1 * t_2 = \dfrac{c}{a}$ from Vieta's formulas. Is this reasoning correct? $\endgroup$ – user592938 Oct 21 '19 at 21:39

Hint: You have the right idea. The solutions of the original equation are then $x=\pm\sqrt{t}$ for each solution $t$ of the quadratic in $t$. These are all real precisely when the roots $t$ are all nonnegative.


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