Solve-the-equation-x-4-2x-3-4x-2-6x-21-0-Given-that-the-sum-of-two-of-its-roots-is-zero-

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 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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