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Question 8

Let $$a,b$$ and $$c$$ be real numbers such that $$2a^2-bc-9a+10=0$$ and $$4b^2+c^2+bc-7a-8=0$$. Then the set of real values that $$a$$ can take is given by

Rewrite the second equation as $$(2b-c)^2+5bc-7a-8=0$$. Adding this to five times the first equation gives $$10a^2+(2b-c)^2-52a+42=0$$. Therefore $$10a^2-52a+42\le0$$, which factors as $$10(a-1)(a-4.2)\le0$$. Hence $$a\in[1,4.2]$$.

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