If the set of all solutions of |x^2 + x - 9| = |x| + |x^2 - 9| is [α, β] ∪ [γ, ∞), then (α^2 + β^2 + γ^2) is equal to:

The given equation is $|x^2 + x - 9| = |x| + |x^2 - 9|$∣x2+x−9∣=∣x∣+∣x2−9∣.
Let $A = x$A=x and $B = x^2 - 9$B=x2−9. Then $A + B = x^2 + x - 9$A+B=x2+x−9.
The equation is of the form $|A + B| = |A| + |B|$∣A+B∣=∣A∣+∣B∣, which holds true if and only if $A \cdot B \ge 0$A⋅B≥0.
Substituting the values of $A$A and $B$B:
$x(x^2 - 9) \ge 0$x(x2−9)≥0
$x(x - 3)(x + 3) \ge 0$x(x−3)(x+3)≥0
The critical points are $-3, 0, 3$−3,0,3. Using the wavy curve method, the solution set is:
$x \in [-3, 0] \cup [3, \infty)$x∈[−3,0]∪[3,∞)
Comparing this with the given solution set $[\alpha, \beta] \cup [\gamma, \infty)$[α,β]∪[γ,∞), we get:
$\alpha = -3$α=−3, $\beta = 0$β=0, $\gamma = 3$γ=3
We need to find the value of $\alpha^2 + \beta^2 + \gamma^2$α2+β2+γ2:
$\alpha^2 + \beta^2 + \gamma^2 = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$α2+β2+γ2=(−3)2+02+32=9+0+9=18
Answer: $18$18