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Algebra Difficulty 4.4 AIME Prove it China

Let A=[−2,4)A = [-2, 4), B={x∣x2−ax−4≤0}B = \{x \mid x^2 - a x - 4 \leq 0\}. If B⊆AB \subseteq A, then the range of real aa is ( ).

Solution

x2−ax−4=0x^2 - a x - 4 = 0 has two roots:
x1=a2−4+a24,x2=a2+4+a24 x_1 = \frac{a}{2} - \sqrt{4 + \frac{a^2}{4}}, \quad x_2 = \frac{a}{2} + \sqrt{4 + \frac{a^2}{4}}
We have B⊆A⇔x1≥−2B \subseteq A \Leftrightarrow x_1 \geq -2 and x2<4x_2 < 4. This means that
a2−4+a24≥−2,a2+4+a24<4 \frac{a}{2} - \sqrt{4 + \frac{a^2}{4}} \geq -2, \quad \frac{a}{2} + \sqrt{4 + \frac{a^2}{4}} < 4
From the above we get 0≤a<30 \leq a < 3.

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