0
$\begingroup$

Algebraically, can't you solve this equation by squaring both sides?: $$\sqrt x = -a $$ $$(\sqrt x)^2 = (-a)^2 $$ $$ x = a ^2$$

I can understand why $x^2 = -a$ has no solution, because a number multiplied by itself twice can't be a negative number. However, in the case above, $-4$ is technically still a square root of $16$.

$\endgroup$
3
  • $\begingroup$ "-4 is techinally still a square root of 16" is incorrect. $\sqrt{b^2}=|b|$ $\endgroup$
    – DatBoi
    Nov 26, 2020 at 6:18
  • 2
    $\begingroup$ @RudyGoburt You need to provide us with a definition of $\sqrt x$. $\endgroup$
    – user838035
    Nov 26, 2020 at 6:20
  • 2
    $\begingroup$ Don't confuse the two square roots of a number, $\pm\sqrt a$, and the square root function $\sqrt a$. $\endgroup$
    – user65203
    Nov 26, 2020 at 7:39

3 Answers 3

4
$\begingroup$

Usually when we write $\sqrt{x}$ where $x$ is a real number, we usually mean the principal root. This means the positive square root. While $(-4)^2=16$, we don't say that $\sqrt{16}=-4$, because $\sqrt{16}=4$, and if $\sqrt{16}$ were both values, it wouldn't be a function. This is why when squaring an equation can sometimes lead to extraneous solutions.

When taking the square root of an equation, we attach a $\pm$. So if $a^2=16$, we do $a=\pm\sqrt{16}=4,-4$.

$\endgroup$
0
$\begingroup$

In the reals, if $a>0$ then $\sqrt x=-a$ has no solution since the square root in this context is usually understood to return a non-negative number.

In the complexes, $\sqrt x=-a$ has two solutions as expected: $x=\pm i\sqrt a$.

$\endgroup$
2
  • $\begingroup$ So it's because the positive solution is used more frequently? $\endgroup$
    – RudyGoburt
    Nov 26, 2020 at 6:24
  • $\begingroup$ @RudyGoburt Non-negative solution is understood by convention. $\endgroup$ Nov 26, 2020 at 6:24
0
$\begingroup$

One more way to see why your argument is not valid,

Assuming $a$ is positive, Define the function $f: [0,\infty) \to [0, \infty)$ by $f(x) = \sqrt x$ and generally when we use $\sqrt x$ we mean this function $f$.

So $\sqrt x$ is a unique,positive value. By definition of $\sqrt x$, it cannot take negative values, hence the equation provided by you does not have a solution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .