Show that the equation $x^4-8x^3+22x^2-24x+4=0$ has exactly two real roots, both being positive There was a question asked on this site which was closed due to lack of showing his attempt. The question was

Show that the equation $$x^4-8x^3+22x^2-24x+4=0$$
has exactly two real roots, both being positive.

I have a solution, it's mine, but I would like an alternative and easy approach to this question.

My approach$:$
$$f(x)=x^4-8x^3+22x^2-24x+4$$
Now $f(x)$ has $4$ sign changes $\implies$ that the number of positive real roots are either $0$ or $2$. But then we see $f(-x)$. There are $0$ sign changes in $f(-x)$ therefore it is $0$ negative real roots. Now we are confirm that if real roots exist they are positive.
Now we calculate the discriminant of $f(x)$. The discriminant of a quartic polynomial is given by this. Calculating it yields the discriminant as $$-25,600$$ Since, this is negative $\implies$ $f(x)$ has two distinct real roots and two imaginary roots.
Since we already proved that the real roots, if exist, have to be positive therefore we proved that the original equation has exactly two real roots, being positive

But can we do this question without involving the concept of discriminant, because it's non-trivial to consider the discriminant of a quartic polynomial because of it's gigantic size. I doubt that anyone can remember it's discriminant that's why I'm asking for an alternative approach, in case it comes in an exam or something like that.
Any help is greatly appreciated.
 A: Note that
$$0=x^4-8x^3+22x^2-24x+4=(x^2-4x+3)^2-5.$$
So $x^2-4x+3=\sqrt 5$ or $x^2-4x+3=-\sqrt 5$.
But $x^2-4x+3=(x-2)^2-1\geq -1>-\sqrt 5$ for $x\in\mathbb R$, thus $x^2-4x+3=\sqrt 5$. Hence $(x-2)^2=\sqrt 5+1$.
Since $\sqrt 5+1<2^2$, the real solutions of $(x-2)^2=\sqrt 5+1$ are $$x_1=2+\sqrt{\sqrt 5+1}>0,\qquad x_2=2-\sqrt{\sqrt 5+1}>0.$$
A: For this particular polynomial it is relatively easy to figure out local extrema:
$$
f^\prime (x) = 4x^3 - 24x^2 + 44x-24 = 4 (x^3 - 6x^2 + 11 x - 6)
$$
$$
f^\prime (1) = 0
\Longrightarrow f^\prime (x) = (x-1) g(x) = 4(x-1) (x^2 - 5x + 6) = 4 (x-1)(x-2)(x-3)
$$
Now we know that the local extrema of $f(x)$ are at points $x=1$, $x=2$ and $x=3$.  But $f(1) = -5$, $f(2) = -4$ and $f(3) = -5$. Since the function values at the local extrema are all negative (and thus the values between those extrema), but the values for large $|x|$ are positive we can show that there are only two roots, i.e. one for $x < 1$ and another for $x > 3$
A: Sometimes it can be helpful to temporarily "remove" the constant term.  The altered function $ \ g(x) \ = \  x^4 - 8x^3 + 22x^2 - 24x  \ $  has one evident zero at $ \ x \ = \ 0 \ $ and testing rational zero candidates of $ \ \  x^3 - 8x^2 + 22x  -24   \ $ shows us that $ \ g(4) \ = \ 0 \ \ , \ $ giving us the factorization $ \ g(x) \ = \ x·(x - 4)·(x^2 - 4x + 6) \ \ , \ $ the third factor being a quadratic polynomial "irreducible over $ \ \mathbb{R}  $ ". Since the leading coefficient is positive, the curve for $ \ g(x) \ $ "opens upward", so there is some sort of "turning point" activity between these zeroes where $ \ g(x) \ < \ 0 \ \ . \ $  We should like to assess where the minimum of this function is.
If we translate $ \ g(x) \ $ "to the left" by $ \ 2 \ $ units, we obtain $$ \ g(x + 2) \ \ = \ \ h(x) \ \ = \ \ (x + 2)·(x-2)·( \ [x + 2]^2 - 4·[x + 2] + 6 \ ) $$ $$ = \ \ (x + 2)·(x-2)·( \ x^2 + 4x + 4 - 4x - 8 + 6 ) \ \ = \ \ (x + 2)·(x-2)·(   x^2 + 2 )   $$ $$ \text{or} \ \ \ x^4 \ - \ 2x^2 \ - \ 8  \ \ . $$
So $ \ h(x) \ $ is symmetrical about the $ \ y-$axis, with only the two real zeroes $ \ x \ = \ \pm 2 \ $ and a $ \ y-$intercept of $ \ -8 \ \ . \ $
(Further probing through the use of calculus indicates that since $ \ h'(x) \ = \ 4x^3   -   4x $ $ = \ 4·x·(x+1)·(x-1)  \ \ $ and $ \ h''(x) \ = \ 12x^2 - 4 \ = \ 4·(\sqrt3 · x + 1)·(\sqrt3 · x - 1) \ \ , \ $ the aforementioned "turning point activity" is a $ \ w-$shaped "wiggle" with relative (and absolute) minima at $ \ ( \pm1 \ , \ -9) \ $ and a relative maximum at $ \ (0 \ , \ -8) \ \ . \ $  This is actually more detail than we'll need.)
Reversing the transformations we've made to the original function, we see even without calculus, that $ \ g(x) \ $ has a "low point" at $ \ (2 \ , \ -8) \ $ (with calculus, we also know that there are minima at $ \ (1 \ , \ -9 ) \ $ and $ \ (3 \ , \ -9) \ $ ) and consequently, that $ \ f(x) \ = \ g(x) + 4 \ \ $ has its "turnaround" in the vicinity of $ \ (2 \ , \ -4) \ \ . \ $  So there can only be two real zeroes, found in the interval $ \ (0 \ , \ 4) \ \ . $
