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Para resolver las ecuaciones cuadráticas de la forma [tex]\(ax^2 + bx + c = 0\)[/tex] utilizando la fórmula general, seguiremos los siguientes pasos:
La fórmula general para resolver una ecuación cuadrática es:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Donde [tex]\(a\)[/tex], [tex]\(b\)[/tex] y [tex]\(c\)[/tex] son los coeficientes de la ecuación cuadrática [tex]\(ax^2 + bx + c = 0\)[/tex].
### a. [tex]\(3x^2 - 14x - 5 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 3\)[/tex], [tex]\(b = -14\)[/tex], [tex]\(c = -5\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-14)^2 - 4 \cdot 3 \cdot (-5) = 196 + 60 = 256 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-(-14) \pm \sqrt{256}}{2 \cdot 3} = \frac{14 \pm 16}{6} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{14 + 16}{6} = \frac{30}{6} = 5 \][/tex]
[tex]\[ x_2 = \frac{14 - 16}{6} = \frac{-2}{6} = -\frac{1}{3} \][/tex]
Entonces, las soluciones son [tex]\(x_1 = 5\)[/tex] y [tex]\(x_2 = -\frac{1}{3}\)[/tex].
### b. [tex]\(2x^2 + 9x + 4 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 2\)[/tex], [tex]\(b = 9\)[/tex], [tex]\(c = 4\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = 9^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2} = \frac{-9 \pm 7}{4} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{-9 + 7}{4} = \frac{-2}{4} = -0.5 \][/tex]
[tex]\[ x_2 = \frac{-9 - 7}{4} = \frac{-16}{4} = -4 \][/tex]
Entonces, las soluciones son [tex]\(x_1 = -0.5\)[/tex] y [tex]\(x_2 = -4\)[/tex].
### c. [tex]\(10x^2 - 13x - 3 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 10\)[/tex], [tex]\(b = -13\)[/tex], [tex]\(c = -3\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-13)^2 - 4 \cdot 10 \cdot (-3) = 169 + 120 = 289 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-(-13) \pm \sqrt{289}}{2 \cdot 10} = \frac{13 \pm 17}{20} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{13 + 17}{20} = \frac{30}{20} = 1.5 \][/tex]
[tex]\[ x_2 = \frac{13 - 17}{20} = \frac{-4}{20} = -0.2 \][/tex]
Entonces, las soluciones son [tex]\(x_1 = 1.5\)[/tex] y [tex]\(x_2 = -0.2\)[/tex].
### d. [tex]\(6x^2 - x - 2 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 6\)[/tex], [tex]\(b = -1\)[/tex], [tex]\(c = -2\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-1)^2 - 4 \cdot 6 \cdot (-2) = 1 + 48 = 49 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-(-1) \pm \sqrt{49}}{2 \cdot 6} = \frac{1 \pm 7}{12} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3} \][/tex]
[tex]\[ x_2 = \frac{1 - 7}{12} = \frac{-6}{12} = -0.5 \][/tex]
Entonces, las soluciones son [tex]\(x_1 = \frac{2}{3}\)[/tex] (o [tex]\(0.666666\)[/tex]) y [tex]\(x_2 = -0.5\)[/tex].
Por lo tanto, las soluciones a las ecuaciones cuadráticas son:
a. [tex]\(x_1 = 5\)[/tex], [tex]\(x_2 = -0.3333333\)[/tex]
b. [tex]\(x_1 = -0.5\)[/tex], [tex]\(x_2 = -4\)[/tex]
c. [tex]\(x_1 = 1.5\)[/tex], [tex]\(x_2 = -0.2\)[/tex]
d. [tex]\(x_1 = 0.666666\)[/tex], [tex]\(x_2 = -0.5\)[/tex]
La fórmula general para resolver una ecuación cuadrática es:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Donde [tex]\(a\)[/tex], [tex]\(b\)[/tex] y [tex]\(c\)[/tex] son los coeficientes de la ecuación cuadrática [tex]\(ax^2 + bx + c = 0\)[/tex].
### a. [tex]\(3x^2 - 14x - 5 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 3\)[/tex], [tex]\(b = -14\)[/tex], [tex]\(c = -5\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-14)^2 - 4 \cdot 3 \cdot (-5) = 196 + 60 = 256 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-(-14) \pm \sqrt{256}}{2 \cdot 3} = \frac{14 \pm 16}{6} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{14 + 16}{6} = \frac{30}{6} = 5 \][/tex]
[tex]\[ x_2 = \frac{14 - 16}{6} = \frac{-2}{6} = -\frac{1}{3} \][/tex]
Entonces, las soluciones son [tex]\(x_1 = 5\)[/tex] y [tex]\(x_2 = -\frac{1}{3}\)[/tex].
### b. [tex]\(2x^2 + 9x + 4 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 2\)[/tex], [tex]\(b = 9\)[/tex], [tex]\(c = 4\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = 9^2 - 4 \cdot 2 \cdot 4 = 81 - 32 = 49 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-9 \pm \sqrt{49}}{2 \cdot 2} = \frac{-9 \pm 7}{4} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{-9 + 7}{4} = \frac{-2}{4} = -0.5 \][/tex]
[tex]\[ x_2 = \frac{-9 - 7}{4} = \frac{-16}{4} = -4 \][/tex]
Entonces, las soluciones son [tex]\(x_1 = -0.5\)[/tex] y [tex]\(x_2 = -4\)[/tex].
### c. [tex]\(10x^2 - 13x - 3 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 10\)[/tex], [tex]\(b = -13\)[/tex], [tex]\(c = -3\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-13)^2 - 4 \cdot 10 \cdot (-3) = 169 + 120 = 289 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-(-13) \pm \sqrt{289}}{2 \cdot 10} = \frac{13 \pm 17}{20} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{13 + 17}{20} = \frac{30}{20} = 1.5 \][/tex]
[tex]\[ x_2 = \frac{13 - 17}{20} = \frac{-4}{20} = -0.2 \][/tex]
Entonces, las soluciones son [tex]\(x_1 = 1.5\)[/tex] y [tex]\(x_2 = -0.2\)[/tex].
### d. [tex]\(6x^2 - x - 2 = 0\)[/tex]
Paso 1: Identificar los coeficientes: [tex]\(a = 6\)[/tex], [tex]\(b = -1\)[/tex], [tex]\(c = -2\)[/tex].
Paso 2: Calcular el discriminante ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac = (-1)^2 - 4 \cdot 6 \cdot (-2) = 1 + 48 = 49 \][/tex]
Paso 3: Sustituir en la fórmula general:
[tex]\[ x = \frac{-(-1) \pm \sqrt{49}}{2 \cdot 6} = \frac{1 \pm 7}{12} \][/tex]
Paso 4: Resolver para los dos valores de [tex]\(x\)[/tex]:
[tex]\[ x_1 = \frac{1 + 7}{12} = \frac{8}{12} = \frac{2}{3} \][/tex]
[tex]\[ x_2 = \frac{1 - 7}{12} = \frac{-6}{12} = -0.5 \][/tex]
Entonces, las soluciones son [tex]\(x_1 = \frac{2}{3}\)[/tex] (o [tex]\(0.666666\)[/tex]) y [tex]\(x_2 = -0.5\)[/tex].
Por lo tanto, las soluciones a las ecuaciones cuadráticas son:
a. [tex]\(x_1 = 5\)[/tex], [tex]\(x_2 = -0.3333333\)[/tex]
b. [tex]\(x_1 = -0.5\)[/tex], [tex]\(x_2 = -4\)[/tex]
c. [tex]\(x_1 = 1.5\)[/tex], [tex]\(x_2 = -0.2\)[/tex]
d. [tex]\(x_1 = 0.666666\)[/tex], [tex]\(x_2 = -0.5\)[/tex]
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