I have this equation:
$$\sqrt{5 - x} = 5 - x^2$$
My current approach is - I note that if I will let: $f(x) = \sqrt{5 - x}, g(x) = 5 - x^2$ then I will have $f(g(x)) = g(f(x)) = x$ Or, in other words, $f(x) = g^{-1}(x)$ (they're inverse) which means that if they intersect, then they must do so on the line y = x . This in turn means that the original equation is same as:
$$x = \sqrt{5 - x} = 5 - x^2$$
Which is of course much easier to solve. Since we have 5 - x under a square root we note that x should not be greater than 5, but also since square root is non-negative, right side should be non-negative as well, thus |x| cannot be greater than $\sqrt{5}$. With this, we can go and solve the quadratic equation:
$x^2 + x - 5 = 0$ this gives two solutions, $x = \frac{-1\pm\sqrt{21}}{2}$
and only one satisfies the condition for $|x|\le\sqrt{5}$ so we conclude $x = \frac{-1+\sqrt{21}}{2}$
Done deal! But.. if I verify this, it turns out this answer is not complete. Take a look:
Clearly there should be one more solution for this. I also plotted $h(x) = -\sqrt{5 - x}$ because that will be the inverse for the negative half of the $g(x) = 5 - x^2$ (the solution to the quadratic which we discarded earlier is the one which solves $-\sqrt{5 - x} = 5 - x^2$ which is also confirmed by $y=x$ passing through that point).
My questions:
- Where's my mistake?
- How to make this work with inverse functions (if possible at all)?
There are two issues in your argument:
First note that the equation $\sqrt{5 - x} = 5 - x^2$ is equivalent to $$\left\{\begin{array}{l}|x|\leq 5\\(x^2-5)^2+x-5=0\end{array}\right.$$ As you noted, every root of $x^2+x-5$ is a root of $(x^2-5)^2+x-5$. Consequently, the polynomial $(x^2-5)^2+x-5$ is divisible by $x^2+x-5$, indeed we have $$(x^2-5)^2+x-5=(x^2+x-5)(x^2-x-4)$$ Hence the third solution of $\sqrt{5 - x} = 5 - x^2$ is a root of $x^2-x-4$.