[Math] Why can we square both sides when LHS and RHS is square rooted

algebra-precalculusarithmetic

When we have something in this form

$$\sqrt{x + a} = \sqrt{y + b},$$

a common technique to solve is to square both side so that:

$$(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b.$$

I'm an engineer and not a mathematician. As I understand it engineers do lots of things that are mathematically frowned upon. However, while those things do make intuitive sense to me this technique makes no sense to me. Specifically what if we consider the alternative road.

\begin{align*}
\sqrt{x + a} &= \sqrt{y + b}\\
\sqrt{x + a} \cdot \sqrt{y + b} &= y + b\\
&\,\, \vdots
\end{align*}

It is unclear to me how we can conclude that $(\sqrt{x + a})^2 = (\sqrt{y + b})^2 \implies x + a = y + b$ is true.

Best Answer

Travelling down the alternative road, we see that

\begin{align*} \sqrt{x+a} &= \sqrt{y+b} & \\ \sqrt{x+a}\times\sqrt{y+b} &= \sqrt{y+b}\times\sqrt{y+b} &(\text{multiplying both sides by}\ \sqrt{y+b})\\ \sqrt{x+a}\times\sqrt{y+b} &= y + b &(\text{simplifying})\\ \sqrt{x+a}\times\sqrt{x+a} &= y + b &(\text{using the equation}\ \sqrt{x+a}=\sqrt{y+b})\\ x + a &= y + b &(\text{simplifying}). \end{align*}

Multiplying by $\sqrt{x+a}$ has exactly the same effect on the equation as multiplying by $\sqrt{y+b}$, but we can replace one expression by the other whenever it is convenient. For example, going from the third equation to the fourth, $\sqrt{y+b}$ on the left hand side was replaced with $\sqrt{x+a}$; this was done to allow for algebraic simplification.

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