To solve the equation:

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

we need to analyze the terms involving absolute values.

1. Analyzing $|x|$:

   • Since $|x| \geq 0$, we can denote $|x| = t$ where $t \geq 0$.
   • This rewrites the equation as $t^2 - 4t + |a - 4| = 0$.

2. Treating $t$ as a Variable:

   • The equation $t^2 - 4t + |a - 4| = 0$ is a quadratic equation in terms of $t$.
   • For this quadratic equation to have real solutions, the discriminant must be non-negative.

3. Calculating the Discriminant:

   • The discriminant $\Delta$ of $t^2 - 4t + |a - 4| = 0$ is given by: $\Delta = (-4)^2 - 4 \cdot 1 \cdot |a - 4| = 16 - 4|a - 4|$
   • For real solutions, $\Delta \geq 0$: $16 - 4|a - 4| \geq 0$ $|a - 4| \leq 4$
   • Solving for $a$, this inequality becomes: $-4 \leq a - 4 \leq 4$ $0 \leq a \leq 8$

4. Solution for $t$:

   • When $0 \leq a \leq 8$, the quadratic equation has real solutions for $t$, which correspond to $|x|$.
   • Solving $t^2 - 4t + |a - 4| = 0$ using the quadratic formula: $t = \frac{4 \pm \sqrt{16 - 4|a - 4|}}{2} = 2 \pm \sqrt{4 - |a - 4|}$

5. Possible Values for $x$:

   • Since $t = |x|$, the solutions for $x$ are: $x = \pm \left(2 \pm \sqrt{4 - |a - 4|}\right)$

Summary of Solutions

The equation $x^2 - 4|x| + |a - 4| = 0$ has real solutions if $0 \leq a \leq 8$. For values of $a$ within this range, the solutions for $x$ are:

$x = \pm \left(2 \pm \sqrt{4 - |a - 4|}\right)$

Would you like a step-by-step explanation of any part, or do you have specific questions?

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Related Questions

1. What happens to the solutions if $a > 8$?
2. How does the solution change if there were no absolute values involved?
3. Can we interpret $a$ geometrically based on the solution conditions?
4. What is the behavior of $x$ as $a$ approaches 0 or 8?
5. How does adding a constant to the equation affect the range of $a$?

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Tip: Always check if conditions involving absolute values yield multiple cases in an equation to avoid missing solutions.