If the set of all solutions of is , then is equal to : [2026]

Question

If the set of all solutions of $| x^{2} + x - 9 | = | x | + | x^{2} - 9 |$ is $[α, β] \cup [γ, \infty)$ , then $(α^{2} + β^{2} + γ^{2})$ is equal to : [2026]

Smart Achievers · Accepted Answer

(2)

We have, $| x^{2} + x - 9 | = | x | + | x^{2} - 9 |$

If $| a + b | = | a | + | b |$ , then $a \cdot b \geq 0$

$∴ x \cdot (x^{2} - 9) \geq 0$

$\Rightarrow x (x + 3) (x - 3) \geq 0$

$\Rightarrow x \in [- 3, 0] \cup [3, \infty)$

Hence, $α = - 3, β = 0, γ = 3$

Thus, $α^{2} + β^{2} + γ^{2} = 18$

Solution

Q.
If the set of all solutions of $| x^{2} + x - 9 | = | x | + | x^{2} - 9 |$ is $[α, β] \cup [γ, \infty)$ , then $(α^{2} + β^{2} + γ^{2})$ is equal to : [2026]

1 9

2 18

3 36

4 72

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