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?
 A: 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 $m \in [\frac{7}{2},\infty)$.
A: One approach is to express $m$ as a function of $x$,
$$m(x)=\frac{x^4+x^2-5}{2x^2-4}
=\frac12\left(x^2+3+\frac1{x^2-2}\right)$$
Then, set $m’(x)=0$ to get 
$$x(x^2-1)(x^2-3)=0$$
which identifies the local extrema at $x=0, \pm 1,\>\pm \sqrt3$. It is straightforward to verify that $m(x)$ has the local minima $m(\pm \sqrt3)=\frac72$ and local maxima $m(0)<m(\pm 1)=\frac32$. (See the plot below.)
Thus, the values of $m$ for real $x$ are
$$m\in (-\infty, \frac32]\cup  [\frac72, \infty)$$
Note the above result assumes that the equation has real roots, but not necessarily real for all roots. There may be some subtlety in interpreting the problem. If all four roots are expected to be real, the lower bound of $m$ is $m(0)=\frac54$ and its values would be $m\in (\frac54, \frac32]\cup  [\frac72, \infty)$. For $m\in (-\infty, \frac54]$, the equation has only one pair of real roots.

A: Put m=2. Then u is not real, so x is not real. Instead, if $ m \ge 5/4 $ write out the two quadratic factors and use the condition for them(it's the same condition for each factor) to have real roots.After simpplification, you finally get $$ (2m-7)(2m-3) \ge  0 $$ so
$$ 1.25 \le m \le 1.5 $$ or
$$ m \ge 3.5 $$
