How do I find this root? How can the third root of $\sqrt{x} - 2x + x^a$ be expressed in terms of a when $a>1.5$? (Obviously there is always one at 1 and 0)
 A: Beside the trivial $x=0$ and $x=1$, the third solution of the equation
$$\sqrt{x} - 2x + x^a=0$$ is close to $\frac 14$ as soon as $a >3$.
So, we can have approximations using Taylor series. For example
$$0=4^{-a}+\left(4^{1-a} a-1\right) \left(x-\frac{1}{4}\right)-2^{-2 a} \left(-8 a^2+8
   a+4^{a}\right)
   \left(x-\frac{1}{4}\right)^2+O\left(\left(x-\frac{1}{4}\right)^3\right)$$ which is a quadratic in $\left(x-\frac{1}{4}\right)$.
Using only the first term (this would be equivalent to the first iterate of Newton method
$$x_1=\frac{1}{4}+\frac{1}{4^a-4 a}$$ is quite decent. For $a=3$, it would give $x=\frac 7{26}=0.269231$ while the "exact" solution is $0.269143$.
But we can continue with series reversion and have
$$x_2=\frac 14+\frac{1}{4^a-4 a}+\frac{8 (a-1) a-4^a}{\left(4^a-4 a\right)^3}+\cdots$$
For $a=3$, this gives $x=\frac{2365}{8788}=0.269117$
Another one which is not bad
$$x_3=\frac 14+\frac{4^a-4 a}{8 a \left(a-4^a+1\right)+4^a+4^{2a}}$$
For the case where $a$ is close to $1.5^+$, the third solution is close to $1$ but, using the same method, it can be approximated as
$$x=1-\frac{4(2a-3)}{4 (a-1) a-1}-\frac{8 (2 a-3)^3 \left(4 a^2-6 a-1\right)}{3 (4 (a-1) a-1)^3}-$$
$$\frac{4 (2 a-3)^3(320 a^6-1728 a^5+3440 a^4-2784 a^3+452 a^2+300 a+27 )}{9 (4 (a-1) a-1)^5}$$
For example, for $a=2$, the above formula gives $x=\frac{6639}{16807}=0.395014$ while the exact solution is $x=\frac{3-\sqrt{5}}{2} =0.381966$.
