# Squaring in a multivariable equation

I'm doing the following exercise:

"If $x = \frac{y}{\sqrt{1 + y^2}}$, then express $y$ in terms of $x$."

I know that we can solve it by squaring and doing the usual arithmetic operations until we get:

$$y^2 = \frac{x^2}{1 - x^2}$$

Then taking the square root to get a final answer of:

$$y = \pm \frac{x}{\sqrt{1 - x^2}}$$

This is the answer given in my textbook. However, I have some doubts because the first step, squaring both sides of the equation, is a non-reversible operation. There could be a solution to the squared equation where $x > 0$ and $y < 0$, which clearly wouldn't be a solution to the original equation.

In equations with a single variable, we can sub our solutions into the original equation and see if they are correct, but what does it mean to square an equation when it contains two or more variables?

• By the way; in this case, it is not possible that $x>0$ and $y<0$ because if $y<0$ then the LHS is $<0$. – Ovi Jun 10 '18 at 5:30

Your doubts are well-founded. The solutions of the squared equation are merely possible solutions of the original equations.

Fortunately, you can check the answer by plugging it into the original equation. First, recall that the $\pm$ sign means there are two proposed solutions. Let’s consider the $+$ solution first: $$y = \frac{x}{\sqrt{1 - x^2}}. \tag1$$ Note that this solution implies $-1<x<1.$ Now we try substituting the expression on the right of Equation $(1)$ for $y$ in the original equation. The result is $$x = \frac{ \left( \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{1+ \left( \frac{x}{\sqrt{1 - x^2}} \right)^2 }} = \frac{ \left( \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{\frac{1}{1 - x^2} }} = x.$$ That is, the substitution results in an equation that is always correct, $x=x.$

Now let’s consider the $-$ solution: $$y = - \frac{x}{\sqrt{1 - x^2}}. \tag2$$

When we try substituting the expression on the right of Equation $(2)$ for $y$ in the original equation, the result is $$x = \frac{ \left( - \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{1+ \left( - \frac{x}{\sqrt{1 - x^2}} \right)^2 }} = -\frac{ \left( \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{\frac{1}{1 - x^2} }} = -x.$$ This is not correct unless $x=0.$ In fact, in the only case in which Equation $(2)$ can be true (the case $x=0,$ $y=0$), Equation $(1)$ is true. The conclusion is that Equation $(1)$ (the $+$ case) is the complete solution, and the $\pm$ in the book’s solution is incorrect.

Another way to approach this question is to observe that $\sqrt{1 + y^2}$ is always positive in the original equation, and therefore $x$ and $y$ always have the same sign. In the proposed solution, $\sqrt{1 - x^2}$ must also be positive. The $+$ solution then ensures that $x$ and $y$ have the same sign, which is required. But the $-$ part of the $\pm$ sign, insofar as it provides an alternative “solution,” says that $x$ and $y$ have opposite signs, which is impossible, except when $x=0,$ in which case the $\pm$ sign has no effect.

At this point you could still do the substitution to verify that the $+$ solution is correct, but there would be no need to check the $-$ alternative.

• I'm still a bit confused, I thought that $\frac{\frac{x}{\sqrt {1 - x^2}}}{\sqrt{\frac{1}{1 - x^2}}}$ could evaluate to $\pm x$ (since the square roots can be either negative or positive). – Pig Jun 10 '18 at 11:36
• Sorry, didn't finish that comment. Second try: I'm still a bit confused, I thought that $\frac{\frac{x}{\sqrt {1 - x^2}}}{\sqrt{\frac{1}{1 - x^2}}}$ could evaluate to $\pm x$ (since the square roots can be either negative or positive). Same for the negative case. As for the second explanation, why must $\sqrt{1 + y^2}$ always be positive? Couldn't we have that $x > 0$, $y < 0$ and $\sqrt{1 + y^2} < 0$? – Pig Jun 10 '18 at 11:49
• $\sqrt v$ means the non-negative square root of $v$ for any non-negative $v.$ It’s that simple. If $\sqrt v$ could be negative then there would be no need to write $\pm \sqrt v$ in formulas. – David K Jun 10 '18 at 19:04
• Okay, thanks, I was confused when I read somewhere that the square root maps an input to two outputs. – Pig Jun 22 '18 at 14:14
• There's the operation of taking a square root of $v,$ which could be interpreted as solving for the value of $x$ in $x^2 = v$ (two solutions may exist), and then there's the symbol $\sqrt{\cdot},$ which according to the usual usage is a function (one output value). So when I say "$\sqrt v$ is the non-negative square root of $v$," implicitly I admit that when $v > 0$ you can say there is also a negative square root of $v$, which would be written $-\sqrt v.$ – David K Jun 22 '18 at 17:50

For the sets of solutions to be preserved you need equivalence relations ($\Leftrightarrow$) between the expressions. Whenever you square you get something called an implication which is not the same as an equivalence.

If you only have implication ($\Leftarrow$ or $\Rightarrow$) then the set of solutions can grow or shrink depending on the direction of the implication.

For example $x = 1 \Rightarrow x^2 = 1^2$

The solution set to the left hand side is $x\in\{1\}$ and to the right one $x\in\{-1,1\}$ and so we see that all elements of left hand solution set are contained in right hand solution set (but not the other way around). This means this algebraic manipulation risks introducing solutions which are not valid solutions to the first equation.

You can compare to famous set theoretic examples. All men are mammals and Adam is a man, therefore Adam must be a mammal. But if we know Lassie is a mammal we can't for sure say he's a man.

If for example you plug in $y= \pm \dfrac 34$ you get $x= \pm \dfrac 35$ and in total there are four correct cases possible as shown below; we can choose the sign in each case,i.e., from each quadrant depending on quadrant context.
• But $y=-3/4$, $x=3/5$ doesn't satisfy the original equation. – Gerry Myerson Jun 9 '18 at 8:51
• Did I understand correctly? When evaluating $x$ for given $y$ the fraction has a plus or minus sign in the denominator due to sqrt sign in $\sqrt {1+y^2}$ from which positive sign only needs to be taken. – Narasimham Jun 9 '18 at 9:03
• The . $\sqrt{..}$ admits/includes by implication what appears to be an extraneous or spurious solution. – Narasimham Jun 13 '18 at 18:47