$$\sqrt{\frac{x}{-4.9}} - \frac{x}{340} = 4.68$$

The following is my work so far:

$$\sqrt{\frac{x}{-4.9}} = 4.68 + \frac{x}{340}$$ $$\frac{x}{-4.9} = 21.9 + \frac{x^2}{115600} + \frac{3182.4x}{115600}$$ $$x^2 + 26774.2x + 2531640 = 0$$

Using the quadratic formula, I get:

$$x_1 = -94.9$$ $$x_2 = -26679.3$$

However, the answer in my book only mentions $-94.9$. I checked a couple of online equation solvers also they only mention -94.9. Additionally, plugging in -26679.3 into the original equation does not work - however plugging it into the derived quadratic does. This must mean my derived quadratic equation is incorrect - does anyone have any idea why?

  • $\begingroup$ You solved it right. Anyhow, since you squared both sides, you got potentially more solutions than your equation has. This means that after getting $x_1,x_2$, you need to also check which, if any works. $\endgroup$ – N. S. Sep 30 '12 at 1:43
  • $\begingroup$ @N.S. Thank You. Sorry, but I don't follow your logic. How can a solution that was derived analytically not be a solution to the equation? $\endgroup$ – user43132 Sep 30 '12 at 1:46
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    $\begingroup$ Is because some of your steps (one actually) is not necessarily reversible. When you have an Equation, lets call it EQ1, and you square it to get EQ2, any solution to your initial equation is also a solution to EQ2. But some solutions to EQ2 could actually come from squaring something like $(-2)=2$. Your particular $x$ can lead to a LHS of $-2$, and a RHS of $2$, so is not a solution to your equation, but becomes a solution after you square both sides... $\endgroup$ – N. S. Sep 30 '12 at 2:35

It's because when you squared both sides of your equation, you made the "negative square root" into a possible solution, whereas $\sqrt{}$ always means the positive square root.

  • $\begingroup$ Without squaring it, how could I solve the equation? $\endgroup$ – user43132 Sep 30 '12 at 1:39
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    $\begingroup$ The other way to solve this is to treat it as a quadratic in $\sqrt{x}$ instead of as a quadratic in $x$. However, what you did is perfectly fine. The lesson to learn is that if you have squared both sides at some point while you solve an equation, then you must check that you didn't introduce an extra solution when you did so. $\endgroup$ – user22805 Sep 30 '12 at 1:51
  • $\begingroup$ Ah, that makes much more sense. Thank you for your help. $\endgroup$ – user43132 Sep 30 '12 at 1:55

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