Absolute values often show up in problems involving square roots. That’s because you can’t take the square root of a negative number without introducing imaginary numbers (those involving i = √-1 ).

Example 1: Simplify √x².

This problem looks deceptively simple. Many students would say the answer is x and move on. However, that is only true for positive values of x.

Try x = 2

√x² = x

Left-hand side: √2² = √4 = 2

Right-hand-side: 2

The equality holds for x = 2 (and actually for all values of x that are ≥ 0).

Try x = -5.

√x² = x

Left-hand side: √(-5)² = √25 = 5

Right-hand-side: -5

Since 5 ≠ -5, this equation does not hold for x = -5. Notice that when x < 0, the left hand side is actually equal to -x. Therefore, for all x < 0, √x² = -x

To summarize, for all x ≥ 0, √x² = x, and for all x < 0, √x² = -x.

We can write this as a piecewise function as follows:

$\sqrt{x^2} =

\begin{cases}

x, & \text{if $x≥0$} \\

-x, & \text{if $x \lt 0$}

\end{cases} $

Our assignment was to simplify, and a piecewise function is surely not simpler than √x². However, this piece-wise function is actually the exact definition of absolute value. Therefore, we can write √x² = |x|, and our simplification is complete. This holds true, for any expression in the square root that has a even exponent. Even though x² is positive, it can be product of two negatives ex: (-5) (-5) or two positives (5)(5). Therefore, the absolute value ensures that both cases are addressed in the solution.

Example 2: Solve for x: x²=z+9

x²=z+9

To solve for x, we first take the square root of both sides.

√x²=√z+9

As we discussed earlier, √x² = |x|, so

|x| = √z+9

The equation isn’t quite solved for x yet. To remove the absolute value, we write:

x = ±√z+9, and our work is done.

When working on problems involving square roots, remember to always check the positive and negative cases and be careful that you don’t miss the absolute value.

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