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