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let f(x) = x^3-6x^2+13x-18 have root a,b, and c. find 1/ab+1/bc+1/ac

 Feb 27, 2021
edited by Charger421  Feb 27, 2021
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We see that we need to find $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}$, which simplifies to $\frac{a+b+c}{abc}.$ From here, we can now proceed with Vieta's. By Vieta's theorem (if we assume that the polynomial $f(x) = Ax^3 + Bx^2 + Cx + D$, we know that $$a+b+c = -\frac{B}{A}, $$ $$ab+ bc + ac = \frac{C}{A},$$ $$abc = -\frac{D}{A},$$ where $A=1, B=-6, C=13, D=-18.$ Therefore, we get that $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{-6}{-18} = \boxed{\frac{1}{3}}.$

 Feb 27, 2021

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