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Question Number 122214 by peter frank last updated on 14/Nov/20

Answered by floor(10²Eta[1]) last updated on 14/Nov/20

$$\mathrm{if}\mathrm{f}\left(\mathrm{x}\right)\mathrm{has}\mathrm{a}\mathrm{repeated}\mathrm{root}\mathrm{so}{\mathrm{f}}^{\prime }\left(\mathrm{x}\right)\mathrm{has}\mathrm{that}\mathrm{root}$$$${\mathrm{f}}^{\prime }\left(\mathrm{x}\right)={54\mathrm{x}}^2+6\mathrm{x}-88=0={27\mathrm{x}}^2+3\mathrm{x}-44$$$$\mathrm{x}=\frac{-1\pm \sqrt{1+4.44.3}}{18}=\frac{-1\pm 23}{18}$$$$\mathrm{x}=\frac{11}{9}\mathrm{or}\mathrm{x}=-\frac{4}{3}$$$$\mathrm{checking}\mathrm{the}\mathrm{solutions}\mathrm{on}\mathrm{the}\mathrm{original}$$$$\mathrm{equation}\mathrm{you}\mathrm{find}\mathrm{that}\frac{-4}{3}\mathrm{is}\mathrm{a}\mathrm{root}$$$$\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+\frac{4}{3})}^2(\mathrm{ax}+\mathrm{b})={18\mathrm{x}}^3+{3\mathrm{x}}^2-88\mathrm{x}-80$$$$\Rightarrow ({\mathrm{x}}^2+\frac{8\mathrm{x}}{3}+\frac{16}{9})(18\mathrm{x}+\mathrm{b})={18\mathrm{x}}^3+(48+\mathrm{b}){\mathrm{x}}^2+(32+\frac{8\mathrm{b}}{3})\mathrm{x}+\frac{16\mathrm{b}}{9}$$$$48+\mathrm{b}=3\Rightarrow \mathrm{b}=-45$$$$\mathrm{f}\left(\mathrm{x}\right)={(\mathrm{x}+\frac{4}{3})}^2(18\mathrm{x}-45)$$$$18\mathrm{x}-45=0\Rightarrow \mathrm{x}=\frac{5}{2}$$$$\mathrm{the}\mathrm{roots}\mathrm{are}\mathrm{x}=\frac{-4}{3},\frac{-4}{3}\mathrm{and}\frac{5}{2}$$

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