If the set of all solutions of $\left| x^{2} + x - 9 \right| = |x| + \left| x^{2} - 9 \right|$ is $\

Question

If the set of all solutions of $\left| x^{2} + x - 9 \right| = |x| + \left| x^{2} - 9 \right|$ is $\lbrack\alpha,\beta\rbrack \cup \lbrack\gamma,\infty)$, then $\left( \alpha^{2} + \beta^{2} + \gamma^{2} \right)$ is equal to :

Accepted Answer

Sol. [Key Concept: $|a| + |b| = |a + b|Leftrightarrow\ ab \geq 0$; so solve $x(x^{2} - 9) \geq 0$.]$${|a| + |b| = |a + b| }{\Rightarrow a \cdot b \geq 0 }{\Rightarrow x\left( x^{2} - 9 \right) \geq 0 }{x \in \lbrack - 3,0\rbrack \cup \lbrack 3,\infty\rbrack }{\alpha^{2} + \beta^{2} + \gamma^{2} = 18 }$$

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