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3(x²- 4) = 0
3(x-2)(x+2)=0
(x-2)(x+2)=0
-->x-2=0
x1=2
-->x+2=0
x2= -2
Comprobando:
3(2)²=12
3( -2)²=12
3x² = 12
x² = 12/3
x² = 4
x = + / - √4 Tiene dos soluciones reales
x₁ = √4
x₁ = 2 Porque 2 * 2 = 4
o
x₂ = - √4
x₂ = - 2 Porque (- 2)(- 2) = 4
Comprobación.
Para x₁ = 2
3x² = 12
3(2)² = 12
3(2 * 2) = 12
3 * 4 = 12
12 = 12
Para x₂ = - 2
3x² = 12
3(- 2)² = 12
3(-2 * - 2) = 12
3 * 4 = 12
12 = 12
Solución.
(2 , - 2)