0

For example consider the equation

$x = 3$

$\implies x^2 = 9 $

$\implies x = 3 \;or\; x = -3$

In this case an extraneous solution is introduced i.e. $ x = -3$

Now for this equation

$\sqrt x = 3 $

$\implies x = 9$

In this case, squaring both sides of the equation doesn't introduce an extraneous solution. I want to ask whats the difference between the first case and the second case and why sometimes squaring both sides can lead to an extraneous solution and why somtimes it doesn't.

Mark Chen
  • 19

3 Answers3

1

In the second case, $\sqrt{x}\ge 0$ must hold, which automatically eliminates the extraneous solution for you.

vadim123
  • 82,796
1

The difference really lies in whether the function you're applying is injective. Essentially, a function $f:A\to B$ is called injective if $f(x)=f(y)$ implies that $x=y$ for all $x,y\in A$, i.e. each output only has one corresponding input. Now, the function $f:\mathbb R\to\mathbb R, x\mapsto x^2$ is not injective, because $(-a)^2=a^2$ but $a\neq -a$ for every nonzero real $a$. Therefore, when we have $a^2=b^2$, we cannot conclude that $a=b$, but we can only say that either $a=b$ or $a=-b$. We have introduced an "extraneous solution". But notice that this is only true because we allow all real--positive and negative--values of $a,b$. In particular, the function $g:\mathbb R_{\geq 0}\to\mathbb R,x\mapsto x^2$ is injective, so if we know that $a,b\geq0$ and $a^2=b^2$, then we do in fact know that $a=b$. No extraneous solution is introduced. In other words, the difference lies in that in the first case, we allowed all real values; but in the second case, we restricted to nonnegative real numbers.

Let's look at your example. In the first case, from $x^2=9$ we tried to conclude $x=3$ or $x=-3$. The solution $x=-3$ came from the fact that we did not know $x$ was positive from the equation. But in the second case, from $x=9$ it is in fact true that $\sqrt x$ can only be $3$, because we are restricting to positive values (recall square roots of real numbers are nonnegative). No extraneous solution introduced.

YiFan Tey
  • 17,431
  • 4
  • 28
  • 66
  • Thanks for the answer! A question I have though is that you defined the function g from nonegative reals to reals but cant sqrt(x) also be a complex number x is usually assumed to be any arbitrary real when solving equations with x? – Mark Chen Jan 25 '19 at 03:08
  • No, $x$ is not assumed to be any arbitrary real quantity. Usually, we assume that everything that appears in the question is real, so $\sqrt x$ is real which means $x$ is nonnegative. Of course you can also consider the $f:\mathbb C\to\mathbb C,z\mapsto z^2$ as well, which is similarly not injective. – YiFan Tey Jan 25 '19 at 03:36
0

The difference is this, in the first instance you find two different solutions for the same problem, in the second instance, you find two different problems for the same solution. The extraneous solution is obviously omitted for the second, instead you have the extraneous problem $\sqrt x =-3$

Rhys Hughes
  • 12,842