Let $f(x) = x^2 + x$, for all real $x$. There exist positive integers $m$ and $n$, and distinct nonzero real numbers $y$ and $z$, such that $f(y) = f(z) = m + \sqrt{n}$ and $f(1/y) + f(1/z) = 1/10$ . Compute $100m + n$.
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Where did you find this question? – bfhaha Mar 24 '18 at 07:35
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Hint. Let $a=m+\sqrt{n}>0$. Note that $y$ and $z$ are the two solutions of the quadratic equation $x^2+x-a=0$. Therefore $y+z=-1$ and $yz=-a$. Hence $$\frac{1}{y}+\frac{1}{z}=\frac{y+z}{yz}=\frac{-1}{-a}=\frac{1}{a}.$$ Moreover $$\frac{1}{y^2}+\frac{1}{z^2}=\frac{y^2+z^2}{y^2z^2}=\frac{(y+z)^2-2yz}{(yz)^2}=\frac{(-1)^2-2(-a)}{(-a)^2}=\frac{1}{a^2}+\frac{2}{a}.$$ Can you take it from here?
Robert Z
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@AnkushYadav So what is $m + \sqrt{n}$? In order to avoid downvotes, show your work! – Robert Z Mar 24 '18 at 08:50
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It is given that f(1/y) + f(1/z) = 1/10 and f(1/y) + f(1/z) = 1/y^2 + 1/y + 1/z^2 + 1/z. After rearranging f(1/y) + f(1/z) = 1/y + 1/z + 1/y^2 + 1/z^2 and according to your result it gives 1/a +2/a + 1/a^2 = 1/10. Therefore, 3/a + 1/a^2 = 1/10. This implies that 10(3a + 1) = a^2. Therefore a^2 - 30a - 10 = 0, and this is a quadratic equation which I solved using quadratic formula and it gives a = 15 + ✓215 or 15 - ✓215. But the second answer will be incorrect. Therefore m + ✓n = 15 + ✓215. – Ankush Yadav Mar 24 '18 at 09:13
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@AnkushYadav Please compute again the solutions of the quadratic equation – Robert Z Mar 24 '18 at 09:17
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I had computed wrong. I think after computing again that the solution is 15 + ✓240. Thanks. – Ankush Yadav Mar 24 '18 at 09:20
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