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HMMT February

Algebra Difficulty 4.8 AIME Find the answer

Find the sum of the xx-coordinates of the distinct points of intersection of the plane curves given by x2=x+y+4x^{2}=x+y+4 and y2=y−15x+36y^{2}=y-15 x+36.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Substituting y=x2−x−4y=x^{2}-x-4 into the second equation yields 0=(x2−x−4)2−(x2−x−4)+15x−36=x4−2x3−7x2+8x+16−x2+x+4+15x−36=x4−2x3−8x2+24x−16=(x−2)(x3−8x+8)=(x−2)2(x2+2x−4)\begin{aligned} 0 & =\left(x^{2}-x-4\right)^{2}-\left(x^{2}-x-4\right)+15 x-36 \\ & =x^{4}-2 x^{3}-7 x^{2}+8 x+16-x^{2}+x+4+15 x-36 \\ & =x^{4}-2 x^{3}-8 x^{2}+24 x-16 \\ & =(x-2)\left(x^{3}-8 x+8\right)=(x-2)^{2}\left(x^{2}+2 x-4\right) \end{aligned} This quartic has three distinct real roots at x=2,−1±5x=2,-1 \pm \sqrt{5}. Each of these yields a distinct point of intersection, so the answer is their sum, 0.

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