An algebra problem by Yellow Tomato

X must be over 3 since under the root is (X-4) => X: 4-9 And since both sides are under squares, and X is positive, then X-1 = 4(sqrt(X-4)) Which mean X-1 should be divisible by 4, which is only satisfied using 5 & 9 for X Substituting gives 5 as result

Given, $\begin{aligned} (x - 1)^{2} & = (4\sqrt{x - 4})^{2} \\ x^{2} - 2x + 1 & = 16x - 64 \\ x^2 - 18x + 65 & = 0 \\ (x - 5)(x - 13) & = 0 \end{aligned}$ Either, $x = 5$ or $x = 13$ . Since $x$ has $1$ digit, $x = \boxed{5}$ is the answer.