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I am trying to learn, formally speaking, what is the correct way of taking the square root of the two sides of an equation in order to solve for the variable. For the discussion, suppose we have to solve

$$ (x-1)^2 = 16 \tag{1} $$ with the existence domain of $x$ being $x\in \mathbb R.$


Method A: setting the argument of the square to be equal to the $\pm$ of rhs.

(I deem this method to be wrong, but I am not sure.)

\begin{align} (1): \sqrt{(x-1)^2} =& \sqrt{16} \\ \Leftrightarrow x-1=-4 \quad \vee& \quad x-1 =4\\ \Leftrightarrow x=-3 \quad \vee& \quad x=5 \end{align}

Although the solutions are right, I think this method is wrong because it does not ensure the square root of the clearly nonnegative $(x-1)^2,$ i.e. $(x-1),$ is nonnegative.


Method B: ensuring the nonnegativity of the square root with absolute values

\begin{align} (1): \sqrt{(x-1)^2} =& \sqrt{16} \\ \Leftrightarrow |x-1| =& 4 \tag{2} \end{align} Removing the absolute values: $$ |x-1|=\begin{cases} x-1 &\text{ if } x\ge 1 \\ -x+1 &\text{ if } x<1. \end{cases} $$ With the piecewise definition, $(2)$ becomes: \begin{align} &|x-1| = 4 \\ &\Leftrightarrow x-1=4 \quad \vee \quad -x+1=4 \\ &\Leftrightarrow x=5 \quad \vee \quad x=-3. \end{align}

This is what I've learned to be the correct way of taking the square root of an equation, ensuring there are no sign ambiguities.


Method C: factorizing using the difference of two squares identity

\begin{align} (1): &\Leftrightarrow (x-1)^2 -16 = 0 \\ &\Leftrightarrow (x-1-4)(x-1+4) = 0\\ &\Leftrightarrow x=5 \quad \vee \quad x=-3. \end{align}

In this method, we forgo having to take the square root thanks to the used identity, so there are no ambiguities to resolve in the process.


Questions:

  1. Am I right in treating Method A as formally imprecise? or is it really equivalent to Method B?
  2. If Method A is indeed wrong, then how do we justify the more commonly used method for solving equations of type \begin{align} x^2&=4 \\ x &= \pm 2 \end{align} without ever specifying the intermediate step invoking the absolute values?
  3. Am I right in considering Method C being correct and devoid of any sign ambiguities?

Admittedly, these are naive questions, but honestly, I am really confused as to what the right way of going about such equations is anymore. So any input would be much appreciated.

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    $\begingroup$ Methods A and B are equivalent as $|x| = y$ if and only if $x = \pm y$. $\endgroup$
    – Zanzag
    Oct 27, 2021 at 20:04
  • $\begingroup$ Methods B and C are the best way to look at square roots, because C is just algebraic manipulation and B follows from the simpel formula $\sqrt{x^2} = |x|$, that works for all $x$, positive or negative $\endgroup$ Oct 27, 2021 at 21:17
  • $\begingroup$ In Method B, by expanding "$|x-1|=4$" to "$x-1=4\;\vee\;-x+1=4$", you're saying exactly "$\pm(x-1)=4$", which is equivalent to saying "$x-1=\pm 4$". (Why?) So, the approach is not "ensuring there are no sign ambiguities"; it's simply moving the ambiguity to a different part of the equation. $\endgroup$
    – Blue
    Oct 28, 2021 at 8:07

1 Answer 1

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$\text{Method - A}$

This method doesn't seem correct to me. Because, your attempts tells us

$$\sqrt {16}=4 \thinspace\thinspace\thinspace\text{or} \thinspace\thinspace\thinspace\sqrt {16}=-4$$

That is not correct. Because, by definition of square root, we have exactly

$$\sqrt {16}=\sqrt {4^2}=|4|=4$$


$\text{Method - B}$

That is correct:

Since both $(x-1)^2≥0$ and $16≥0$, then we have

$$\begin{align}&(x-1)^2=16\\ \iff &\sqrt{(x-1)^2}=\sqrt {16}\\ \iff &|x-1|=4\\ \iff &\begin{cases}x-1=4,\thinspace \text{if}\thinspace x≥1\\ -x+1=4,\thinspace \text{if} \thinspace x<1\end{cases} \\ \iff &x_1=5\thinspace \thinspace\thinspace\text{and}\thinspace\thinspace \thinspace x_2=-3\end{align}$$

But, you can also write

$$|x-1|=4\iff x-1=±4$$

by definition of absolute value.

Because, we can simply see that if $a≥0$, then $|±a|=a$.


$\text{Method - C}$

Absolutely correct. Because

$$\begin{align}&A^2=B^2\\ \iff &A^2-B^2=0\\ \iff &(A-B)(A+B)=0\\ \iff &A-B=0 \thinspace\thinspace\thinspace\text{or}\thinspace\thinspace\thinspace A+B=0.\end{align}$$

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