Equations with Absolute Values: Quadratic equation

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.

To solve the equation $|x^2 - 6x + 8| = 0$, recognize that the absolute value of a number is zero if and only if the number itself is zero. Therefore, we set the expression inside the absolute value to zero:

$x^2 - 6x + 8 = 0$

Next, we attempt to factor the quadratic expression:

The expression $x^2 - 6x + 8$ can be factored into:

$(x - 2)(x - 4) = 0$

Now, apply the zero product property, which states if a product equals zero, at least one of the factors must be zero. So, set each factor equal to zero:

• $x - 2 = 0$ gives $x = 2$
• $x - 4 = 0$ gives $x = 4$

Thus, the solutions to the equation $|x^2 - 6x + 8| = 0$ are:

$x = 2$ and $x = 4$.

To solve the equation $|x^2 + 4| = 40$, we consider two cases based on the properties of absolute value:

• Case 1: $x^2 + 4 = 40$
  - Subtract 4 from both sides to isolate the quadratic term:
  $x^2 = 40 - 4$
  $x^2 = 36$
  - Solving for $x$, take the square root of both sides:
  $x = \pm \sqrt{36}$
  $x = \pm 6$

• Case 2: $x^2 + 4 = -40$
  - Subtract 4 from both sides:
  $x^2 = -40 - 4$
  $x^2 = -44$
  - Since $x^2 = -44$ has no real number solutions (as the square of a real number cannot be negative), we discard this case.

Therefore, the only valid solutions are from Case 1: $x = \pm 6$.

Thus, the solution to the equation $|x^2 + 4| = 40$ is $x = \pm 6$.

To solve the equation $|x^2 - 11| = 5$, we will consider the two cases implied by the property of absolute values:

• Case 1: $x^2 - 11 = 5$
• Case 2: $x^2 - 11 = -5$

Let's solve each case separately.

Case 1: $x^2 - 11 = 5$
First, add 11 to both sides: $x^2 = 16$.
Taking the square root of both sides, we get $x = \pm\sqrt{16} = \pm4$.
Thus, the solutions for Case 1 are $x = 4$ and $x = -4$.

Case 2: $x^2 - 11 = -5$
First, add 11 to both sides: $x^2 = 6$.
Taking the square root of both sides, we get $x = \pm\sqrt{6}$.
Thus, the solutions for Case 2 are $x = \sqrt{6}$ and $x = -\sqrt{6}$.

Therefore, combining solutions from both cases, the complete set of solutions is $x = 4, -4, \sqrt{6}, -\sqrt{6}$.

Reviewing the multiple-choice options, both cases' solutions correspond to the correct answer listed as 'Answers a + b.'