Question Number 106637 by john santu last updated on 06/Aug/20
$$\:\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\mathrm{The}\:\mathrm{quartic}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{3}} +\mathrm{14x}+\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{one}\:\mathrm{root}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1}+\mathrm{2i}\:.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{other}\:\mathrm{three}\:\mathrm{roots}.\: \\ $$
Answered by bemath last updated on 06/Aug/20
![_(@bemath@) as the coefficient of the quartic are real; it follows that 1−2i is also a root. then [x−(1−2i)][x−(1+2i)] is a factor of the quartic. ⇒x^2 −2x+5 . Therefore x^4 +2x^3 +14x+15=(x^2 −2x+5)(x^2 +ax+b) and we get a = 4 and b = 3. p(x)=(x^2 −2x+5)(x^2 +4x+3)=0 → { ((x_(1,2) =((2±4i)/2) = 1±2i)),((x_(3,4) =((−4±2)/2)=−2±1)) :}](Q106640.png)
$$\:\:\underset{@\mathrm{bemath}@} {\:} \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quartic}\:\mathrm{are}\:\mathrm{real};\:\mathrm{it}\:\mathrm{follows} \\ $$$$\mathrm{that}\:\mathrm{1}−\mathrm{2i}\:\mathrm{is}\:\mathrm{also}\:\mathrm{a}\:\mathrm{root}.\:\mathrm{then}\: \\ $$$$\left[\mathrm{x}−\left(\mathrm{1}−\mathrm{2i}\right)\right]\left[\mathrm{x}−\left(\mathrm{1}+\mathrm{2i}\right)\right]\:\mathrm{is}\:\mathrm{a}\:\mathrm{factor}\: \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{quartic}.\: \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{5}\:.\:\mathrm{Therefore}\: \\ $$$$\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{3}} +\mathrm{14x}+\mathrm{15}=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{5}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{ax}+\mathrm{b}\right) \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:=\:\mathrm{4}\:\mathrm{and}\:\mathrm{b}\:=\:\mathrm{3}. \\ $$$$\mathrm{p}\left(\mathrm{x}\right)=\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{5}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4x}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\rightarrow\begin{cases}{\mathrm{x}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{2}\pm\mathrm{4i}}{\mathrm{2}}\:=\:\mathrm{1}\pm\mathrm{2i}}\\{\mathrm{x}_{\mathrm{3},\mathrm{4}} =\frac{−\mathrm{4}\pm\mathrm{2}}{\mathrm{2}}=−\mathrm{2}\pm\mathrm{1}}\end{cases} \\ $$
Commented by john santu last updated on 06/Aug/20
$$\mathfrak{good}\: \\ $$