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.
Best Answer
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 $$x_1=2+\sqrt{\sqrt 5+1}>0,\qquad x_2=2-\sqrt{\sqrt 5+1}>0.$$