$f(x) = 1 - 6/x + 8a/x^{3/2} - 3a^2/x^2$ where $-1 How do you show that $f(x)$ has a root at $x>1+(1-a^2)^{1/2}$ for all $a$? It can be shown that $f(x)$ has a local minimum at $x=a^2$ at which $f<0$ and that $f(x)$ approaches $1$ for large $x$, so the root must be located at $x>a^2$.
Find root of $f(x) = 1 - 6/x + 8a/x^{3/2} - 3a^2/x^2$
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polynomials
roots
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0My first guess would be to show that $f$ must change sign – 2017-02-06
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0Consider $$x^2-6x+8\sqrt{(1-b^2)x}-3(1-b^2)$$ at $$x=1+b$$ and show it is negative. This reduces to showing $$\sqrt{(1+b)(1-b^2)}<1+\tfrac12b-\tfrac12b^2$$ Squaring both sides, one sees that this holds at every $|b|<1$. – 2017-02-06
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0@Did I think you are right. Thanks! – 2017-02-06