For a software project, I need to calculate some things. on of the formulas looks like:

26280 = 2 * SQRT((149,598,000,000 - x) / 1.4) + x / 10,217,588,910,387,300,000

My colegue says you can't solve the above equation (you won't be able to find x) I quite convinced you should be able to find x, the problem is, I don't know how :(

I came as far as rewriting the above into:

((26280 - x) / 20,435,177,820,774,600,000)^2 = (149,598,000,000 - x) / 1.4

but now I'm stuck.

Could anyone explain to me how to move on, in order to find x?

  • $\begingroup$ Expand everything and use the quadratic equation. $\endgroup$ – Austin Mohr Apr 5 '12 at 13:19
  • $\begingroup$ it's the 'expand everything' that I don't know how to do $\endgroup$ – Flipse Apr 5 '12 at 13:28
  • $\begingroup$ is it true that (a/b)^2 = a^2 / b^2 ? (it's been at least 15 years since I last had math classes, and only 10 years after that I learned that I should have been paying attention) $\endgroup$ – Flipse Apr 5 '12 at 13:33
  • $\begingroup$ Yes. That expression is true. $\endgroup$ – dot dot Apr 5 '12 at 13:36
  • $\begingroup$ The two equations you have written are not equivalent. The two answers below address the second one, but it's not the same as the first. Because the numbers are so large, the solutions to both equations are approximately the same. $\endgroup$ – Peter Phipps Apr 5 '12 at 15:08

First, it's useful to write the expression in the standard quadratic form. For ease of reading/writing, I'm going to set $A = 20,435,177,820,744,600,00$ and $B = 149,598,000,000$.

$$ \begin{align*} \left(\frac{26280 - x}{A}\right)^2 &= \frac{B - x}{1.4}\\ \frac{(26280 - x)^2}{A^2} &= \frac{B}{1.4} - \frac{x}{1.4}\\ \frac{26280^2 - 52560x - x^2}{A^2} &= \frac{5}{7}B - \frac{5}{7}x\\ \frac{26280^2}{A^2} - \frac{52560x}{A^2} - \frac{x^2}{A^2} &= \frac{5}{7}B - \frac{5}{7}x\\ 0 &= \frac{x^2}{A^2} + \left(\frac{52560}{A^2} - \frac{5}{7}\right)x + \frac{5}{7}B - \frac{26280^2}{A^2}. \end{align*} $$

You can finish this up with the quadratic formula.

EDIT: Alternatively, there is WolframAlpha.

  • $\begingroup$ Is it guaranteed that both solutions of the quadratic equation will solve the original equation? $\endgroup$ – Johannes Kloos Apr 5 '12 at 13:39
  • $\begingroup$ @Johannes Kloos: Yes. For example x=1 has the same solution as 2x=2 because the relation within x and the right hand of the equation is the same in both. $\endgroup$ – dot dot Apr 5 '12 at 13:42
  • $\begingroup$ @dotdot: That works only because you are applying an equivalence transformation. As a counter-example, $x=-1$ has one solution, namely $-1$. $x^2 = (-1)^2 = 1$ has two, $x = 1$ and $x=-1$. $\endgroup$ – Johannes Kloos Apr 5 '12 at 13:50
  • $\begingroup$ thanks! that helped me across. $\endgroup$ – Flipse Apr 5 '12 at 13:52
  • $\begingroup$ As Johannes points out, you should make sure the two solutions given by the quadratic are in the domain of your original equation (the one with the square root). $\endgroup$ – Austin Mohr Apr 5 '12 at 14:06

As Austin Mohr said it's a quadratic equation. You must only expand. Their solutions are: -298283208976198672964919958167422500864 and 149598000000.


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