Why doesn't squaring both sides produce extraneous roots here?

Question

$$\sqrt{\frac{\left(x-2\right)^{2}+\left(y-3\right)^{2}}{\left(x+1\right)^{2}+\left(y-4\right)^{2}}}=\frac{2}{3}...(i)$$

$$5x^{2}+5y^{2}-44x-22y+49=0...(ii)$$

(i) is the original equation, and (ii) is the simplified version of (i), which I have found by squaring both sides of (i). Even though (ii) is the squared version of (i), the graphs of the two equations are identical. Why is this the case? Why doesn't (ii) have extraneous values?

This might help you in answering the question.

Answer 1

Generally, $a(x)=b(x)$ is equivalent to $a(x)-b(x)=0$, while $a^2(x)=b^2(x)$ is equivalent to

$$0=a^2(x)-b^2(x)=(a(x)-b(x)(a(x)+b(x)).$$

The last equation shows where extraneous solution come from when "squaring": they are solutions of $a(x)+b(x) = 0$.

It just so happens that in your case, where $a(x)=\sqrt{\ldots}$ and $b(x)=\frac23$, which are both non-negative (and the second actually positive), so their sum can never be zero!

In addition, in practical work with such equations, extraneous solutions also come from non-rigorously defined (or simply not computed) domains for the variable to be calculated.

For example, the equation

$$\sqrt{x+1} = \sqrt{x^2+2x-1}$$

has a single solution in $\mathbb R$, namely $x=2$. The squared equation

$$x+1 = x^2+2x-1$$

however has 2 real solutions: $x_1=2, x_2=-1$. The extraneous solution $x_2=-1$ comes from the fact that the original equation is meaningless for that $x$ value (when considering real numbers only). The original equation only makes sense (when working with real numbers) when $x \in [\sqrt{2}-1,\infty).$

Confined to that domain, the squared equation has no extraneous solutions.

But in a real world scenario, you often do not bother making a complete analysis of when the original equation is well-defined (which may be a harder problem than solving it!), but you apply squaring and other "non-equivalent" techniques and then have to remember that the solutions you find at the end are just "candidate solutions" for the original equation.

Answer 2

The reason is simple. The equation states that the square root of some quantity equals $2/3$. If you square both sides, it just means that the quantity must equal $4/9$. Nothing has changed, because the only way the LHS can equal $2/3$ is if we take the positive root of the quantity inside the root.

Another way to put it is to consider a simple example: $$\sqrt{z} = 3.$$ In this case, what is the solution set for $z$? It cannot be anything else besides $9$. There is no introduction of an extraneous solution.

Then, why is it that extraneous solution are introduced for, say, $$z = 9 \implies z^2 = 81?$$ In this case, squaring both sides introduces the solution $z = -9$ because the act of squaring lets us choose a negative value, which when squared, becomes positive. But this is not permitted in the square root situation above, since $\sqrt{-9} \ne 3$.

Answer 3

Consider two equations:

$$\sqrt{x+y}=4...(i)$$

$$\sqrt{x+y}+x=4...(ii)$$

Now, in equation (i),

$$\sqrt{x+y}=4$$

$$\iff x+y=16...(iii)$$

Again, in equation (ii),

$$\sqrt{x+y}+x=4$$

$$\implies x+y+2x\sqrt{x+y}+x^{2}=16...(iv)$$

(i) & (iii) are equivalent equations (graphs are identical), but (ii) & (iv) aren't (graphs aren't identical). For this reason, (iii) doesn't have extraneous roots but (iv) does.