# Solving $400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$ for $x$?

I am trying to solve $$400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}} = 0$$ for $$x$$. Thanks to an earlier question I put up here, I know $$x = 6- \frac{2}{\sqrt{3}}$$, but I want to figure out why. Symbolab gives a ridiculously complicated equation for $$x$$, and I couldn't figure it out myself when I tried.

Please help me figure out why $$x = 6 - \frac{2}{\sqrt{3}}$$.

Some of what I tried:

$$400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}}=0$$

Subtract $$400,000$$ from both sides.

$$\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}}=400,000$$

Multiply both sides by $$\sqrt{4+(6-x)^2}$$.

$$800,000(6-x)=400,000\sqrt{4+(6-x)^2}$$

Divide both sides by $$800,000$$.

$$6-x=\frac{\sqrt{4+(6-x)^2}}{2}$$

Multiply both sides by $$2$$.

$$12-2x=\sqrt{4+(6-x)^2}$$

Take both sides to the power of $$2$$.

$$-4x^2-48x+144= -x^2-12x+40$$

• The last step makes no sense...that's not how you divide polynomials. – lulu Apr 12 '19 at 23:21
• 1) before squaring both sides, one should first evaluate the signs. 2) when you squared both sides, some inappropriate minus signs appeared and the $4$ on the RHS disappeared. – Saucy O'Path Apr 12 '19 at 23:22
• Nor is there any sense to the division in the first place. Just combine like terms to see that you are down to a quadratic equation. Keep in mind that you must still test the roots...since you squared you might have picked up some false solutions. – lulu Apr 12 '19 at 23:22
• @SaucyO'Path thank you, I will fix the dissappearing four. What do you mean when you say, "evaluate the signs"? – LuminousNutria Apr 12 '19 at 23:25
• I mean using the equivalence $\sqrt A=B\iff \begin{cases}A\ge 0\\ B\ge 0\\ A=B^2\end{cases}$; at least if you assume the usual $\sqrt\bullet:[0,\infty)\to [0,\infty)$. – Saucy O'Path Apr 12 '19 at 23:26

You don't quite have the correct solution.

\begin{align*} 400\,000 - \frac{800\,000 (6-x)}{\sqrt{4+(6-x)^2}} &= 0 \\ - \frac{800\,000 (6-x)}{\sqrt{4+(6-x)^2}} &= -400\,000 \\ \frac{2 (6-x)}{\sqrt{4+(6-x)^2}} &= 1 \\ 2 (6-x) = \sqrt{4+(6-x)^2} &\text{ or } \sqrt{4+(6-x)^2} = 0 \text{,} \end{align*} because if you inadvertently multiply both sides of your equation by zero, you get a true equation ... that tells you nothing about your variables. On the right, the only number whose square root is $$0$$ is $$0$$, so $$4 + (6-x)^2 = 0$$, which is impossible, since no real number squares to $$-4$$. So we continue on the left. \begin{align*} 2 (6-x) &= \sqrt{4+(6-x)^2} \\ 4 (6-x)^2 &= 4+(6-x)^2 \end{align*} Since squaring can introduce spurious solutions, we must check all our eventual solutions back in the original equation. \begin{align*} 3 (6-x)^2 &= 4 \\ (6-x)^2 &= 4/3 \\ 6-x &= 2/\sqrt{3} \\ x-6 &= -2/\sqrt{3} \\ x &= 6 - 2/\sqrt{3} \text{.} \end{align*} Verifying that this solution is not spurious:\begin{align*} &400\,000 - \frac{800\,000 (6-(6 - 2/\sqrt{3}))}{\sqrt{4+(6-(6 - 2/\sqrt{3}))^2}} \\ &\quad = 400\,000 - \frac{800\,000 (2/\sqrt{3})}{\sqrt{4+(2/\sqrt{3})^2}} \\ &\quad = 400\,000 - \frac{800\,000 (2/\sqrt{3})}{\sqrt{4+(2/\sqrt{3})^2}} \\ &\quad = 400\,000 - \frac{800\,000 (2/\sqrt{3})}{\sqrt{4+4/3}} \\ &\quad = 400\,000 - \frac{800\,000 (2/\sqrt{3})}{\sqrt{16/3}} \\ &\quad = 400\,000 - \frac{800\,000 (2/\sqrt{3})}{4/\sqrt{3}} \\ &\quad = 400\,000 - 800\,000 (2/\sqrt{3}) \cdot \frac{\sqrt{3}}{4} \\ &\quad = 400\,000 - 800\,000 \cdot \frac{2}{4} \\ &\quad = 400\,000 - 400\,000 \\ &\quad = 0 \text{,} \end{align*} so the one solution found is not spurious.

• Thank you, but how did you go from $4 (6-x)^2 = 4+(6-x)^2$ to $3 (6-x)^2 = 4$? – LuminousNutria Apr 12 '19 at 23:44
• @LuminousNutria Subtract $(6-x)^2$ from both sides. – Toby Mak Apr 13 '19 at 0:36
• @TobyMak Thank you – LuminousNutria Apr 13 '19 at 0:39

The exercise does not have to be so tedious.

$$400,000-\frac{800,000(6-x)}{\sqrt{4+(6-x)^2}}=0\tag{1}$$

First, divide by $$400,000$$'

$$1-\frac{2(6-x)}{\sqrt{4+(6-x)^2}}=0$$

Substitute $$u=6-x$$

$$1-\frac{2u}{\sqrt{4+u^2}}=0$$

Multiply by $$\sqrt{4+u^2}$$.

$$\sqrt{4+u^2}-2u=0$$

$$\sqrt{4+u^2}=2u\tag{2}$$

Square both sides and solve for $$u$$.

$$\begin{eqnarray} 4+u^2&=&4u^2\\ 3u^2&=&4\\ u&=&\pm\frac{2}{\sqrt{3}} \end{eqnarray}$$

Check for extraneous solutions by checking the results in equation (2).

The negative solution for $$u$$ is extraneous, however

$$\sqrt{4+\left(\frac{2}{\sqrt{3}}\right)^2}=\sqrt{4+\frac{4}{3}}=\frac{4}{\sqrt{3}}=2\cdot\frac{2}{\sqrt{3}}=2u$$

So $$u=6-x=\frac{2}{\sqrt{3}}$$. Thus $$x=6-\frac{2}{\sqrt{3}}$$

• This is useful. Thank you! – LuminousNutria Apr 14 '19 at 17:32