To solve the equation $|4 - x^2| = 2 - x$, we need to consider two cases for the absolute value, since $|A| = B$ implies that $A = B$ or $A = -B$. So, we will solve for both cases:

Case 1: $4 - x^2 = 2 - x$

In this case, we can solve the equation directly:

$4 - x^2 = 2 - x$ Rearrange the equation:

$x^2 - x + 2 - 4 = 0$

$x^2 - x - 2 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}$

$x = \frac{1 \pm \sqrt{1 + 8}}{2}$

$x = \frac{1 \pm \sqrt{9}}{2}$

$x = \frac{1 \pm 3}{2}$

So, the two possible solutions are:

$x = \frac{1 + 3}{2} = 2 \quad \text{or} \quad x = \frac{1 - 3}{2} = -1$

Case 2: $4 - x^2 = -(2 - x)$

Now, solve for the other case where $4 - x^2 = -(2 - x)$:

$4 - x^2 = -2 + x$

Rearrange the terms:

$x^2 - x + 2 + 4 = 0$

$x^2 - x + 6 = 0$

We solve this using the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(6)}}{2(1)}$

$x = \frac{1 \pm \sqrt{1 - 24}}{2}$

$x = \frac{1 \pm \sqrt{-23}}{2}$

Since the discriminant is negative, there are no real solutions for this case.

Final Solutions:

From Case 1, the real solutions are $x = 2$ and $x = -1$. We should check which of these satisfies the original equation $|4 - x^2| = 2 - x$:

• For $x = 2$: $|4 - (2)^2| = |4 - 4| = 0 \quad \text{and} \quad 2 - 2 = 0$ This is true.

• For $x = -1$: $|4 - (-1)^2| = |4 - 1| = 3 \quad \text{and} \quad 2 - (-1) = 3$ This is true.

Therefore, the solutions are $x = 2$ and $x = -1$.

Final Answer:

$x = 2 \quad \text{or} \quad x = -1$

Would you like any additional details or have any questions?

Here are some related questions:

1. How do you solve absolute value equations in general?
2. What are the conditions for quadratic equations to have real solutions?
3. How does the quadratic formula work, and why is the discriminant important?
4. What are the steps to graph an absolute value function?
5. Can absolute value equations have no real solutions?

Tip: Always check both cases when solving absolute value equations!