$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers

Question

$$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$

My attempt

$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$

$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$

$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$

$2x + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$

$x + \sqrt{x^2 + 2(938^2 + 140^2)x+(938*140)^2} = 1116^2 - 938^2 - 140^2$

At this point trying to solve for x inside the sqrt in the quadratic gives me an imaginary number. How is it possible to solve this?

Answer 1

For this sort of problem, sometimes it will be useful to replace explicit numbers by variables to avoid distractions.

Let $a = 938$, $b = 140$, $c = 938 + 140 + 38 = 1116$.

Let $u = \sqrt{x+a^2}$ and $v = \sqrt{x+b^2}$, the equation at hand becomes

$$u + v = c\tag{*1}$$

Notice $u^2 - v^2 = (x+a^2) - (x+b^2) = a^2-b^2$, we have

$$u - v = \frac{u^2 - v^2}{u+v} = \frac{a^2-b^2}{c}\tag{*2}$$

Combine $(*1)$ and $(*2)$, we have $\displaystyle\;u = \frac{c^2 + a^2-b^2}{2c}$ and hence $$\begin{align} x = u^2 - a^2 & = \left(\frac{c^2 + a^2-b^2}{2c}\right)^2 - a^2 = \left(\frac{(c+a)^2 - b^2}{2c}\right)\left(\frac{(c-a)^2-b^2}{2c}\right)\\ & = \frac{(c+a+b)(c+a-b)(c-a+b)(c-a-b)}{4c^2}\\ & = \frac{2194 \cdot 1914 \cdot 318 \cdot 38}{4\cdot 1116^2} = \frac{352392601}{34596} \approx 10185.93481905423 \end{align} $$

Answer 2

Repeated squaring is not necessary. Let $y > -140$ be such that $x = y^2 + 280y$, so that $$\sqrt{x+140^2} - 140 = \sqrt{(y + 140)^2} - 140 = y.$$ Then $$\begin{align*} 38 &= \sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 \\ &= \sqrt{y^2 + 280y + 938^2} - 938 + y, \end{align*}$$ or equivalently, $$976-y = \sqrt{y^2 + 280y + 938^2}.$$ Now we square only once, giving $$y^2 - 2(976)y + 976^2 = y^2 + 280y + 938^2.$$ The quadratic terms cancel, and we get $$2(140+976)y = 976^2 - 938^2 = (976-938)(976+938) = 38(1914),$$ or $$y = \frac{6061}{186}.$$ Then substitute back to find the solution for $x$.

Answer 3

You received good answers and solutions. However, it seems that you made a few mistakes in your calculations. The line $$2x + 2\sqrt{x+938^2}\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$$ is correct. What is not correct is in the next line since $$(x+938^2)(x + 140^2)=x^2+(938^2+140^2)x+938^2\times 140^2$$ I do not know how appeared the factor $2$ in the radical and disappeared in other places.

Using this correction, we then have $$2x+2\sqrt{x^2+(938^2+140^2)x+938^2\times 140^2}=346012$$ Dividing both sides by $2$ and moving the $x$ term to the rhs then gives $$\sqrt{x^2+899444 x+17244942400}=173006-x$$ Squaring again, the $x^2$ disappear and what is left is $$1245456 x-12686133636=0\implies x=\frac{352392601}{34596}$$

But, as Achille Hui answered, replace your numbers by letters to avoid these enormous intermediate numbers.

Answer 4

After the first line of your attempt, multiply both sides by $\sqrt{x+938^2}-\sqrt{x+140^2}$

$$x+938^2-(x+140^2)=1116(\sqrt{x+938^2}-\sqrt{x+140^2})$$

Divide both sides by $1116$.

$$\dfrac{938^2-140^2}{1116}=\sqrt{x+938^2}-\sqrt{x+140^2}$$

Taken with the previous equation

$$\sqrt{x+938^2}+\sqrt{x+140^2}=1116$$

we can add both equations to solve for $\sqrt{x+938^2}$ or subtract to solve for $\sqrt{x+140^2}$. Then simply square both sides and solve for $x$.