Find solutions of $F(x)=F^{-1}(x)$

Question

$F:(-\infty,\frac{5}{4})\rightarrow(\frac{-9}{8},\infty)$

$F(x) =2x^2-5x+2$ , we need to find solutions of $F(x)=F^{-1}(x)$

Obviously one solution can be find $F(x)=x$ such that x<$\frac{5}{4}$, which is $\frac{3-\sqrt{5}}{2}$

For other solution i thought (a,b) and (b,a) should be solution of $F(x)$ , which is not a very good method as i had to solve biquadratic which was solved by dumb luck .(2 roots were 0)

So unoredered pair of 'a' and 'b' come out to be (0,2) (deleting $F(x)=x$) which is not possible as both should be less than $\frac{5}{4}$.

So my doubt is, is there a better way to solve $F(x)=F^{-1}(x)$(expect $F(x)=x$). Or do u have to rely on above method, finding roots by hit and trial and using that fact the equation formed also satisfies roots of $F(x)=x$(i think directly equating $F(x)$ =$F^{-1}(x)$ also gives same equations , pls tell me if i am wrong here also).

Answer 1

Let $y=2x^2-5x+2.$ Thus, $x=2y^2-5y+2,$ which gives $$y-x=2(x-y)(x+y)-5(x-y)$$ or $$(x-y)(x+y-2)=0.$$ Can you end it now?

Answer 2

Some general insight could be attempted here with aid of graphs.

Taking a function and its inverse $F(x),F^{-1}(x)$ and solving for them together $F(x)-F^{-1}(x)=0, $ we get points of intersection say $(x1,y1),(x2,y2).$

1) There is a line of mirror symmetry $ x-y=0, $ through origin and one intersection point. (Blue)

2) There is also a straight line perpendicular to the symmetry line passing through the other two intersection points. (Brown)

$$ \dfrac{y-y1}{x-x1}= \dfrac{y1-y2}{x1-x2} $$

(This condition appears true for such second degree curves only).

These we can verify by taking general parabola and its inverse with axes parallel and perpendicular to x-axis ( Red and Green ) respectively. When solving these together we get

$$ y-( ax^2+bx+c)=0,\, x-(ay^2+by+c)=0 $$

and after factoring

$$ x-y=0,\, x+y=\frac{-(1+b)}{a}; $$

In the particular case here $ (a,b,c)=(2,-5,2)$ and intersection points are $(0,2),(2,0).$