1. Solving for x

$\sqrt{1+\sqrt{x-1}} + \sqrt{4-\sqrt{x-1}} = 3$

$A = \sqrt{x-1}$

$1 + A + 2\sqrt{(1+A)(4-A)} + 4 - A = 3$

$2\sqrt{(1+A)(4-A)} = -2$

$\sqrt{(1+A)(4-A)} = -1$

$(1+A)(4-A) = 1$

$4 +3A - A^{2} = 1$

$3A - A^{2} = -3$

$A^{2} -3A = 3$

$x - 1 -3\sqrt{x-1} = 3$

$x -3\sqrt{x-1} = 4$

$-3\sqrt{x-1} = 4 - x$

$9(x-1) = x^{2} - 8x + 16$

$x^2 - 17x + 25 = 0$

And now I'm stuck. What did I do wrong?

2. From $A^2 - 3A = 3$

$A^2 - 3A + \left(-\frac{3}{2}\right)^2 = 3 + \left(-\frac{3}{2}\right)^2$

$\left(A - \frac{3}{2}\right)^2 = 3 + \frac{9}{4}$

$\left(A - \frac{3}{2}\right)^2 = \frac{21}{4}$

$A - \frac{3}{2} = \pm \frac{\sqrt{21}}{2}$

$A = \frac{3}{2} \pm \frac{\sqrt{21}}{2}$.

Therefore $\sqrt{x - 1} = \frac{3}{2} \pm \frac{\sqrt{21}}{2}$

$x - 1 = \left(\frac{3}{2} \pm \frac{\sqrt{21}}{2}\right)^2$

$x - 1 = \frac{9}{4} \pm \frac{3\sqrt{21}}{2} + \frac{21}{4}$

$x - 1 = \frac{15}{2} \pm \frac{3\sqrt{21}}{2}$

$x = \frac{17}{2} \pm \frac{3\sqrt{21}}{2}$.

Therefore the two solutions are

$x = \frac{17 - 3\sqrt{21}}{2}$ and $x = \frac{17 + 3\sqrt{21}}{2}$.

3. Sorry for the hassle. I just discovered that I forgot to do 3^2 in the right sight of the second equation.