Solve the equation x^4 − 2x^3 + 4x^2 + 6x − 21 = 0, Given that the sum of two of its roots is zero


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Question Number 192370 by Tawa11 last updated on 15/May/23

$Solve the equation x^{4} - {2 x}^{3} + {4 x}^{2} + 6 x - 21 = 0,$ $Given that the sum of two of its roots is zero$
	Answered by Frix last updated on 16/May/23

	$(x - a) (x + a) (x - b) (x - c) = 0$ $x^{4} - (b + c) x^{3} - (a^{2} - b c) x^{3} + a^{2} (b + c) x - a^{2} b c = 0$ $[1] b + c = 2$ $[2] a^{2} - b c = - 4$ $[3] a^{2} (b + c) = 6$ $[4] a^{2} b c = 21$ $[1] & [3] \Rightarrow a^{2} = 3 \Rightarrow a = \pm \sqrt{3}$ $\Rightarrow$ $[2], [4] b c = 7$ $\Rightarrow$ $b + c = 2 \land b c = 7$ $\Rightarrow b = 1 \pm \sqrt{6} i \land c = 1 \mp \sqrt{6} i$
	Commented by Tawa11 last updated on 16/May/23

	$God bless you sir .$
	Answered by a.lgnaoui last updated on 16/May/23
	$l equatoon admetant deux racines opposees (α et −α verifie[alors l equation (x^2 −𝛂)(x^2 +ax+b) x^4 +ax^3 + (b− 𝛂)x^2 − aα x− bα=0 ⇒ a=−2 b−𝛂 =4 ∗ −a𝛂 =+6 ⇒ α=((+6)/2) = +3 −bα =−21 ⇒ b=((21)/3) = +7 ∗b−α=7−3=4 (verifie) donc l equation (E) (x^2 −3)(x^2 −2x+7) { ((x_1 =(√3) x_2 =−x_1 =−(√3) )),((x_3 =((2−2i(√6))/2) x_4 =((2+2i(√6))/2) (△=−24))) :} [ x_3 =1−i(√(6 )) x_4 =1+i(√6) ]$
	$l equatoon admetant deux r a cines opposees$ $(α et - α verifie [alors l equation$ $(x^{2} - α) (x^{2} + ax + b)$ $x^{4} + {ax}^{3} + (b - α) x^{2} - a α x - b α = 0$ $\Rightarrow a = - 2$ $b - α = 4 $ $- a α = + 6 \Rightarrow α = \frac{+ 6}{2} = + 3$ $- b α = - 21 \Rightarrow b = \frac{21}{3} = + 7$ $ b - α = 7 - 3 = 4 (verifie)$ $donc l equation (E)$ $(x^{2} - 3) (x^{2} - 2 x + 7)$ ${\begin{cases} x_{1} = \sqrt{3} x_{2} = - x_{1} = - \sqrt{3} \\ x_{3} = \frac{2 - 2 i \sqrt{6}}{2} x_{4} = \frac{2 + 2 i \sqrt{6}}{2} (△ = - 24) \end{cases}$ $[x_{3} = 1 - i \sqrt{6} x_{4} = 1 + i \sqrt{6}]$
	Commented by Tawa11 last updated on 16/May/23

	$God bless you sir .$
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