I was solving a Euclidean geometry problem and I got the equation $\sqrt{x^2+20\sqrt2 x – 100} + \sqrt{x^2 + 25} = 15$. I asked Wolfram Alpha and it says the answer is $60\sqrt3-70\sqrt2$. But I'm wondering is it possible to solve it by hand? (without using softwares like Wolfram Alpha)
Before I check the answer with WA, I thought the answer should be in the form $x=a\sqrt2$ and substituted it in the equation. (I justified it as each term on the LHS should be a non-negative integer) not sure why the answer is in the form $x= a\sqrt 2 +b \sqrt 3$.
I also tried squaring both sides of the original equation but it seems that makes things more complicated.
Best Answer
Here's one way to approach a problem like this. First, multiply both sides by $\sqrt{x^2+20\sqrt2 x - 100} - \sqrt{x^2 + 25}$ to get
$$\begin{equation}\begin{aligned} (x^2 + 20\sqrt{2}x - 100) - (x^2 + 25) & = 15\left(\sqrt{x^2 + 20\sqrt{2}x - 100} - \sqrt{x^2 + 25}\right) \\ 20\sqrt{2}x - 125 & = 15\left(\sqrt{x^2 + 20\sqrt{2}x - 100} - \sqrt{x^2 + 25}\right) \\ 4\sqrt{2}x - 25 & = 3\left(\sqrt{x^2 + 20\sqrt{2}x - 100} - \sqrt{x^2 + 25}\right) \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, for simpler algebra, let
$$y = \sqrt{x^2 + 20\sqrt{2}x - 100}, \;\; z = \sqrt{x^2 + 25} \tag{2}\label{eq2A}$$
Thus, the original equation along with the result in \eqref{eq1A} gives the result of
$$y + z = 15, \;\; 3y - 3z = 4\sqrt{2}x - 25 \tag{3}\label{eq3A}$$
Multiplying the first equation above by $3$ and adding it to the second one gives
$$\begin{equation}\begin{aligned} 6y & = 4\sqrt{2}x + 20 \\ 3\sqrt{x^2 + 20\sqrt{2}x - 100} & = 2\sqrt{2}x + 10 \\ 9(x^2 + 20\sqrt{2}x - 100) & = (2\sqrt{2}x + 10)^2 \\ 9x^2 + 180\sqrt{2}x - 900 & = 8x^2 + 40\sqrt{2}x + 100 \\ x^2 + 140\sqrt{2}x - 1000 & = 0 \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Note that multiplying both sides of the first equation in \eqref{eq3A} by $3$ and subtracting it from the second one, to get $6z = -4\sqrt{2}x + 70$ instead, results (after squaring both sides and simplifying) in the same equation as in \eqref{eq4A}. Thus, using either method, there's now just a quadratic equation to solve, and then check both of its roots to determine which one(s) satisfy the original equation.