# Find the possible values of the parameter $m$ for which the given equation has all of its roots real.

I am given the following equation:

$$x^4+(2m-1)x^2+2m+2=0$$

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 $$\boxed{\text{C}}$$.

• I must be forgetting something - why do you need to make sure that both of the roots are nonnegative? – Axion004 Oct 21 '19 at 1:52
• @Axion004 Still need to sub back in $t=x^2$. No real roots for $x$ if $t$ is negative. – Andrew Chin Oct 21 '19 at 1:55
• Also, why does $2m−1\le 0$ for the roots to be nonnegative? – Axion004 Oct 21 '19 at 2:51
• @Axion004 in $a(x-\alpha)(x-\beta)=ax^2+bx+c=0$, we have negative $\alpha+\beta=b$ and positive $\alpha\beta=c$ for $\alpha,\beta>0$. – Andrew Chin Oct 21 '19 at 2:58
• @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? – 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.