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Algebra Difficulty 4.8 AIME Prove it Bulgaria

Problem:
Find all real numbers aa for which the equation x2+ax+3a2−7a−19=0x^{2}+a x+3 a^{2}-7 a-19=0 has real roots x1x_{1} and x2x_{2} such that
1x1−2+1x2−2=−2a13 \frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=-\frac{2 a}{13}

Solution

Solution:
Using Vieta's formulas we get
1x1−2+1x2−2=x1+x2−4(x1−2)(x2−2)=−a+43a2−5a−15 \frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=\frac{x_{1}+x_{2}-4}{\left(x_{1}-2\right)\left(x_{2}-2\right)}=-\frac{a+4}{3 a^{2}-5 a-15}
Therefore 3a2−5a−15≠03 a^{2}-5 a-15 \neq 0 and
a+43a2−5a−15=2a13 \frac{a+4}{3 a^{2}-5 a-15}=\frac{2 a}{13}
Hence 6a3−10a2−43a−52=0⟺(a−4)(6a2+14a+13)=06 a^{3}-10 a^{2}-43 a-52=0 \Longleftrightarrow (a-4)\left(6 a^{2}+14 a+13\right)=0, i.e. a=4a=4. In this case x1,2=−2±15x_{1,2}=-2 \pm \sqrt{15}.

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