If the set of all solutions of |x^2 + x - 9| | vedclass

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(B) We know that $|A + B| = |A| + |B|$ if and only if $AB \geq 0$.Given the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$.Let $A = x$ and $B = x^2 - 9$.

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) We know that $|A + B| = |A| + |B|$ if and only if $AB \geq 0$.Given the equation $|x^2 + x - 9| = |x| + |x^2 - 9|$.Let $A = x$ and $B = x^2 - 9$. Then $A + B = x^2 + x - 9$.The condition for the equality to hold is $x(x^2 - 9) \geq 0$.Factoring the expression,we get $x(x - 3)(x + 3) \geq 0$.Using the wavy curve method (sign scheme),the solution set is $x \in [-3, 0] \cup [3, \infty)$.Comparing this with the given form $[\alpha, \beta] \cup [\gamma, \infty)$,we identify $\alpha = -3$,$\beta = 0$,and $\gamma = 3$.Therefore,the value of $(\alpha^2 + \beta^2 + \gamma^2) = (-3)^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$.

If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$,then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:

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