Fermat 2008 Q22

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Problem Statement

For how many integers $k$ do the parabolas with equations $y = -\dfrac{1}{8}x^2 + 4$ and $y = x^2 - k$ intersect on or above the x-axis?

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Solution

$$\text{let } f(x) = -\dfrac{1}{8}x^2 + 4 \implies \text{ vertex is at } (0, 4) \text{ and the parabola opens down}$$ $$\text{let } g(x) = x^2 - k \implies \text{ vertex is at } (0, -k) \text { and the parabola opens up}$$ $$\implies g(x) \text{ does not intersect with } f(x) \text{ when k < -4} \implies min(k) = -4$$ $$\text{set } f(x) = 0 \implies -\dfrac{1}{8}x^2 + 4 = 0 \implies x^2 - 32 = 0 \implies (x - \sqrt{32})(x + \sqrt{32}) = 0$$ $$\text{Roots } (\sqrt{32}, 0) \text{ and } (-\sqrt{32}, 0)$$ $$\text{set } g(x) = 0 \implies x^2 - k = 0 \implies (x - \sqrt{k})(x + \sqrt{k}) = 0$$ $$\text{Roots } (\sqrt{k}, 0) \text{ and } (-\sqrt{k}, 0)$$ $$\implies \text{if } f(x) \text{ intersects } g(x) \text{ on the x-axis at } (\sqrt{32}, 0) \text{ and } (-\sqrt{32}, 0) \text { when } k = 32$$ $$\implies max(k) = 32$$ $$\therefore -4 <= k <= 32 \implies \exists \; 32 - (-4) + 1 = 37 \text { possible values for } k$$