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rezolvarea ecuatiei x^2+(2-√3)x-2√3=0

Răspuns :

102533

Răspuns:

Explicație pas cu pas:

x² + (2 - √3)·x -2√3 = 0

x₁,₂ = [(√3 - 2)±√(4+3-4√3+8√3)]/2

x₁,₂ = [(√3 - 2) ± √(7+4√3)] /2

√(7+4√3) = √(7+√48) = √[7+√(49-48)/2] +√[7-√(49-48)/2] =

=√(7+1)/2 +√(7-1)/2 = √4 + √3 = 2 + √3

x₁,₂ = [(√3 - 2)±(2+√3)]/2

x₁ = (√3 - 2 + 2 +√3)/2 = √3

x₂ = (√3 - 2 - 2 - √3) /2 = -2

x₁ = √3  ; x₂ = -2

Rayzen

[tex]x^2+(2-\sqrt 3)x-2\sqrt 3 = 0\\ \\ \Delta = (2-\sqrt 3)^2 -4\cdot (-2\sqrt 3) = (2-\sqrt 3)^2+8\sqrt 3 = \\ =4-4\sqrt 3+3+8\sqrt 3 = 7+4\sqrt 3 = (2+\sqrt 3)^2 \\ \\ x_{1,2} = \dfrac{-(2-\sqrt 3)\pm \sqrt{ (2+\sqrt 3)^2}}{2} = \dfrac{\sqrt 3 - 2\pm |2+\sqrt 3|}{2} = \\ \\ = \dfrac{\sqrt 3 - 2 \pm (2+\sqrt 3)}{2} \Rightarrow \left|\begin{array}{lcl}x_1 = \dfrac{\sqrt 3-2-2-\sqrt 3}{2}\Rightarrow \boxed{x_1 = -2}\\ x_2 = \dfrac{\sqrt 3-2+2+\sqrt 3}{2}\Rightarrow \boxed{x_2 = \sqrt 3}\end{array}[/tex]

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