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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:

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

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

If not, how to solve such equation?

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3 Answers

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:

$$a\sqrt{\frac{bc^2}{a^2b}}=a\sqrt{\frac{c^2}{a^2}}=a\left|\frac{c}a\right|\;.\tag{1}$$

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 $bx$.

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$a\sqrt{bx} = c$

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

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

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

This is essentially your argument.

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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 $\sqrt{x} - 1/\sqrt{x} = -2/\sqrt{3}$.

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