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Question Number 106637 by john santu last updated on 06/Aug/20

$$@\mathrm{JS}@$$$$\mathrm{The}\mathrm{quartic}\mathrm{equation}{\mathrm{x}}^4+{2\mathrm{x}}^3+14\mathrm{x}+15=0$$$$\mathrm{has}\mathrm{one}\mathrm{root}\mathrm{equal}\mathrm{to}1+2\mathrm{i}.\mathrm{Find}$$$$\mathrm{the}\mathrm{other}\mathrm{three}\mathrm{roots}.$$

Answered by bemath last updated on 06/Aug/20

$$\underset{@\mathrm{bemath}@}{}$$$$\mathrm{as}\mathrm{the}\mathrm{coefficient}\mathrm{of}\mathrm{the}\mathrm{quartic}\mathrm{are}\mathrm{real};\mathrm{it}\mathrm{follows}$$$$\mathrm{that}1-2\mathrm{i}\mathrm{is}\mathrm{also}\mathrm{a}\mathrm{root}.\mathrm{then}$$$$[\mathrm{x}-(1-2\mathrm{i})][\mathrm{x}-(1+2\mathrm{i})]\mathrm{is}\mathrm{a}\mathrm{factor}$$$$\mathrm{of}\mathrm{the}\mathrm{quartic}.$$$$\Rightarrow {\mathrm{x}}^2-2\mathrm{x}+5.\mathrm{Therefore}$$$${\mathrm{x}}^4+{2\mathrm{x}}^3+14\mathrm{x}+15=({\mathrm{x}}^2-2\mathrm{x}+5)({\mathrm{x}}^2+\mathrm{ax}+\mathrm{b})$$$$\mathrm{and}\mathrm{we}\mathrm{get}\mathrm{a}=4\mathrm{and}\mathrm{b}=3.$$$$\mathrm{p}\left(\mathrm{x}\right)=({\mathrm{x}}^2-2\mathrm{x}+5)({\mathrm{x}}^2+4\mathrm{x}+3)=0$$$$\to \{\begin{array}{l}{\mathrm{x}}_{1,2}=\frac{2\pm 4\mathrm{i}}{2}=1\pm 2\mathrm{i}\\ {\mathrm{x}}_{3,4}=\frac{-4\pm 2}{2}=-2\pm 1\end{array}$$

Commented by john santu last updated on 06/Aug/20

$$\mathfrak{good}$$

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