Solving $\sqrt{x^2+20\sqrt2 x - 100} \space + $ $\sqrt{x^2 + 25} $ $= 15$ by hand

Question

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.

Answer 1

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.

Answer 2

It is possible to do it by hand, though the calculations are not pretty. The radicals can be eliminated by squaring twice, which gives the following (at most) quartic equation.

$$ \left(\left(x^2 + 20 \sqrt{2} x - 100\right) + \left(x^2 + 25\right) - 225\right)^2 = 4 \left(x^2 + 20 \sqrt{2} x - 100\right) \left(x^2 + 25\right) $$

Leaving the intermediate calculations to WA, the final result reduces to the quadratic:

$$ x^2 + 140 \sqrt{2} x - 1000 = 0 $$

The positive root is indeed $60 \sqrt{3} - 70 \sqrt{2}$.

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[ EDIT ] $\;$ Not something to try by hand, but another way to transform the equation into an algebraic one with integer coefficients is to use WA's GroebnerBasis, which yields:

$$ x^4 - 41200 x^2 + 1000000 = 0 $$

Solving the quadratic in $x^2$ then denesting the radicals gives the same root as before.

Answer 3

Squaring both sides will work just fine if you move the $\sqrt{x^2+25}$ to the other side first.

$$\begin{align} \sqrt{x^2 + 20\sqrt2 \, x - 100} &= 15 - \sqrt{x^2 + 25} \\[2mm] x^2 + 20\sqrt2 \, x - 100 &= 225 - 30 \sqrt{x^2 + 25} + (x^2 + 25) \\[2mm] &= x^2 + 250 - 30 \sqrt{x^2 + 25} \\[2mm] 2\sqrt2 \, x - 35 &= 3 \sqrt{x^2 + 25} \\[2mm] 8x^2 -140\sqrt{2} \, x + 1225 &= 9x^2 + 225 \\[2mm] x^2 + 140\sqrt{2} \, x - 1000 &= 0 \end{align}$$

Now just apply the quadratic formula:

$$\begin{align} x &= \frac{-140\sqrt{2} \pm \sqrt{(140\sqrt{2})^2 + 4(1000)}}{2} \\[2mm] &= -70\sqrt{2} \; \pm \ 60\sqrt{3} \end{align}$$

You should be able to verify that $-70\sqrt{2} - 60\sqrt{3} \;$ is an extraneous root.