The problem asks for the value of $(\alpha^2 + \beta^2 + \gamma^2)$ derived from the solution set of the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$, which is given as $[\alpha, \beta] \cup [\gamma, \infty)$.
Understanding the Absolute Value Equation
We utilize a key property of absolute values: $|a| = |b| + |c|$ holds if and only if $a = b + c$ and the product $b \cdot c \ge 0$.
In this equation, let $a = x^2 + x - 9$, $b = x$, and $c = x^2 - 9$. First, we verify if $a = b + c$: $b + c = x + (x^2 - 9) = x^2 + x - 9$ This matches $a$. Therefore, the original equation $|x^2 + x - 9| = |x| + |x^2 - 9|$ is equivalent to the condition that $b$ and $c$ have the same sign or one/both are zero, meaning $b \cdot c \ge 0$.
Solving the Inequality Condition
The condition simplifies to:
$x \cdot (x^2 - 9) \ge 0$Factorizing the quadratic term gives:
$x(x - 3)(x + 3) \ge 0$The critical points where the expression equals zero are $x = -3$, $x = 0$, and $x = 3$. We test the sign of $x(x - 3)(x + 3)$ in the intervals determined by these points:
- Interval $x < -3$: Negative (e.g., $(-4)(-7)(-1) < 0$)
- Interval $-3 \le x \le 0$: Non-negative (e.g., $(-1)(-4)(2) > 0$)
- Interval $0 < x < 3$: Negative (e.g., $(1)(-2)(4) < 0$)
- Interval $x \ge 3$: Non-negative (e.g., $(4)(1)(7) > 0$)
The inequality $x(x - 3)(x + 3) \ge 0$ is satisfied when $-3 \le x \le 0$ or $x \ge 3$.
Identifying Solution Set Parameters
The derived solution set is $[-3, 0] \cup [3, \infty)$.
Comparing this with the given format $[\alpha, \beta] \cup [\gamma, \infty)$, we find:
- $\alpha = -3$
- $\beta = 0$
- $\gamma = 3$
Final Calculation
We now calculate the required value $(\alpha^2 + \beta^2 + \gamma^2)$: $ \alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + (0)^2 + (3)^2 $ $ = 9 + 0 + 9 $ $ = 18 $
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