Question 19: The inequality \(\sqrt {2{x^3} + 3{x^2} + 6x + 16} – \sqrt {4 – x} \ge 2\sqrt 3 \) has a solution set of [a; b]. What is the total value of a2+b2?

Condition: -2 ≤ x ≤ 4.

Consider

on paragraph [ -2; 4].Yes \(f’\left( x \right) = \frac{{3\left( {{x^2} + x + 1} \right)}}{{\sqrt {2{x^3} + 3 {x^2} + 6x + 16} }} + \frac{1}{{2\sqrt {4 – x} }} > 0\;,\forall x \in \left( { – 2;4} \ right).\)

Therefore, the covariate function on [-2; 4]

The given inequality becomes \(f\left( x \right) \ge f\left( 1 \right) = 2\sqrt 3 \)

Combined with the condition that the covariate function infers x ≥ 1.

Compared to the condition, the solution set of bpt is [1; 4].

Therefore; a^{2}+ b^{2 }= 17.

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