Another Radical Equation

Given an equation similar to $\sqrt{x-1}+\sqrt{x+4} = 5$ , we can always let the smallest radical be $\sqrt u = \sqrt {x-1}$ , then the equation becomes $\sqrt u+\sqrt{u+a} = b$ . In this example, $a=b=5$ .

$\begin{aligned} \sqrt u+\sqrt{u+a} & = b \\ \sqrt{u+a} & = b - \sqrt u & \small \color{#3D99F6} \text{Squaring both sides} \\ u + a & = u - 2b \sqrt u + b^2 \\ 2b \sqrt u & = b^2 - a \\ \implies u & = \left(\frac {b^2 - a}{2b}\right)^2 & \small \color{#3D99F6} \text{For }a=b=5 \\ u & = \left( \frac {5^2 - 5}{2\cdot 5}\right)^2 \\ x-1 & = 4 \\ \implies x & = \boxed{5} \end{aligned}$

In general: $\sqrt{ x - 1 } + \sqrt{ x + 4 } = n$ $\sqrt{x+4}=n-\sqrt{x-1}$ Squaring we get: $x+4=n^2+(x-1)-2n\sqrt{x-1}$ $\implies \sqrt{x-1}=\dfrac{n^2-5}{2n}~~,n\ne 0$ Squaring again gives: $\boxed{x=1+\left[\dfrac{n^2-5}{2n}\right]^2}$ .

For $n=5$ , we get $x=5$ . A direct substitution in the original equation confirms the answer.

[This is not yet a solution. If no one adds a non-trial and error solution in a week, I will flesh out the details.]

So, most people might eventually guess that $x = 5$ gives us $\sqrt{ x - 1 } + \sqrt{ x + 4 } = \sqrt{ 4 } + \sqrt{ 9 } = 5$ .

However, this is somewhat haphazard, and doesn't allow us to solve the more general problem. In fact, how can we generalize the problem?

$\begin{matrix} \sqrt{x-1}+\sqrt{x+4}=5\\ \sqrt{x-1}+\sqrt{x+4}-\sqrt{x+4}=5-\sqrt{x+4}\\ \sqrt{x-1}=5-\sqrt{x+4}\\ \left(\sqrt{x-1}\right)^2=\left(5-\sqrt{x+4}\right)\\ x-1=x+29-10\sqrt{x+4}\\ x-1-x=x+29-10\sqrt{x+4}-x\\ -1=-10\sqrt{x+4}+29\\ -1-29=-10\sqrt{x+4}+29-29\\ -30=-10\sqrt{x+4}\\ \left(-30\right)^2=\left(-10\sqrt{x+4}\right)^2\\ 900=100(x+4)\\ 900=100x+400\\ 900-400=100x+400-400\\ 500=100x\\ \frac{500}{100}=\frac{100x}{100}\\ 5=x\\ \textbf{x=5} \end{matrix}$

( √(x-1) + √(x+5) )²=5²
x-1+x+4+2√(x²+3x-4)=25
2x+2√(x²+3x-4)=22
x+√(x²+3x-4) =11
( √(x²+3x-4) )²=(11-x)²
x²+3x-4=121-22x+x²
25x=125
x=5