Solving |x²+4| = 20: An Absolute Value Equation Challenge

To solve the equation $|x^2 + 4| = 20$, we need to consider the definition of absolute value, which gives us two equations to solve:

• First equation: $x^2 + 4 = 20$
• Second equation: $x^2 + 4 = -20$

Let's solve each one individually:

For the first equation, $x^2 + 4 = 20$:

Subtract 4 from both sides to isolate the $x^2$ term:

$x^2 = 20 - 4$

$x^2 = 16$

Taking the square root of both sides gives us the solutions:

$x = \pm \sqrt{16}$

$x = \pm 4$

Now, consider the second equation, $x^2 + 4 = -20$:

Subtract 4 from both sides:

$x^2 = -20 - 4$

$x^2 = -24$

This provides no real solutions since a square cannot be negative in the real number system.

Therefore, the solutions to the original equation are $x = \pm 4$, which corresponds to choice 2.