Find $m$ such that $x^4 - (2m - 1)x^2 + 4m -5 = 0$ has real roots Consider the equation:
$$ x ^ 4 - (2m - 1) x^ 2 + 4m -5 = 0 $$
with $m \in \mathbb{R}$. I have to find the values of $m$ such that the given equation has all of its roots real.
This is what I did:
Let $ u = x^2, \hspace{.25cm} u\ge 0$
We get:
$$ u ^ 2 - (2m - 1)u + 4m -5 = 0 $$
Now since we have 
$$ u = x ^ 2$$
That means
$$x = \pm \sqrt{u}$$
That means that the roots $x$ are real only if $u \ge 0$.
So we need to find the values of $m$ such that all $u$'s are $\ge 0$. If all $u$'s are $\ge 0$, that means that the sum of $u$'s is $\ge 0$ and the product of $u$'s is $ \ge 0 $. Using Vieta's formulas
$$S = u_1 + u_2 = - \dfrac{b}{a} \hspace{2cm} P = u_1 \cdot u_2 = \dfrac{c}{a}$$
where $a, b$ and $c$ are the coefficients of the quadratic, we can solve for $m$. We get:
$$S = - \dfrac{-(2m - 1)}{1} = 2m - 1$$
We need $S \ge 0$, so that means $m \ge \dfrac{1}{2}$ $(1)$
$$P = \dfrac{4m - 5 }{1} = 4m - 5$$
We need $P \ge 0$, so that means $m \ge \dfrac{5}{4}$ $(2)$
Intersecting $(1)$ and $(2)$ we get the final answer:
$$ m \in \bigg [ \dfrac{5}{4}, \infty \bigg )$$
My question is: Is this correct? Is my reasoning sound? Is there another way (maybe even a better way!) to solve this?

You need also to consider $\Delta$ to be positive in order for the solutions to be real.
$\Delta = (2m-1)^2-4(4m-5)=4m^2-4m+1-16m+20=4m^2-20m+21$ 
$m_{1,2}=\frac{10 \pm \sqrt{100-84}}{4}=\frac{10 \pm 4}{4}=\{\frac{3}{2},\frac{7}{2} \}$
Thus $\Delta \geq 0 \Leftrightarrow m \in (-\infty, \frac{3}{2}] \cup [\frac{7}{2},\infty)$
Take for example $m=2$: now $m \geq \frac{5}{4}$ but the equation 
$$
u^2-(2 \cdot 2 -1)u+4\cdot 2-5=0 \\
u^2-3u+3=0
$$
has not real solutions $\frac{3 \pm \sqrt{-3}}{2}$.
Hence we need $m \in [\frac{7}{2},\infty)$.
