Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Sorry if this is a very primitive question, but I really not sure if I am right about this kind of situations. Imagine the following equation where $a$ , $b$ and $c$ are known numbers and $x$ is the unknown variable:


Is it ok in this case to do it like $$a^2bx=c^2$$

If not, how to solve such equation?

share|cite|improve this question
up vote 3 down vote accepted

Yes, this is fine, provided that $a$ and $c$ have the same algebraic sign. When you solve the second equation, you get $$x=\frac{c^2}{a^2b}\;.$$ Now try substituting that into the original equation:


If $a$ and $c$ have the same algebraic sign, $\left|\dfrac{c}a\right|=\dfrac{c}a$, and $(1)$ can be simplified to $a\left(\dfrac{c}a\right)=c$, as desired.

If one of $a$ and $c$ is positive and the other negative, the original equation has no solution, since by convention $\sqrt{bx}$ denotes the non-negative square root of $bx$.

share|cite|improve this answer

$a\sqrt{bx} = c$

$\sqrt{bx} = \frac{c}{a}$

$bx = \frac{c^2}{a^2}$

$x = \frac{c^2}{ba^2}$

This is essentially your argument.

share|cite|improve this answer

In general, you have to be careful to check each "solution" by plugging it in to the original equation: this sort of argument often introduces extraneous roots, because squaring is not a one-to-one function. For example, try $$ \sqrt{x} - 1/\sqrt{x} = 2/\sqrt{3}$$ Squaring both sides and expanding gives you $$ x - 2 + 1/x = 4/3 $$ which has solutions $x=3$ and $x=1/3$. But only $x=3$ is a solution of the original equation: $x=1/3$ is instead a solution of $\sqrt{x} - 1/\sqrt{x} = -2/\sqrt{3}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.