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f) \( x^{2}-7 x+6 \leq 0 \)

Ask by Craig Christensen. in Argentina
Jan 23,2025

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Answer

La solución de la desigualdad \( x^{2} - 7x + 6 \leq 0 \) es \( x \) entre 1 y 6, inclusive.

Solution

Para resolver la desigualdad cuadrática \( x^{2} - 7x + 6 \leq 0 \), seguiremos estos pasos: ### 1. Encontrar las raíces de la ecuación cuadrática Primero, resolvemos la ecuación cuadrática asociada: \[ x^{2} - 7x + 6 = 0 \] Factorizamos el trinomio: \[ x^{2} - 7x + 6 = (x - 1)(x - 6) = 0 \] Por lo tanto, las raíces son: \[ x = 1 \quad \text{y} \quad x = 6 \] ### 2. Determinar los intervalos Las raíces dividen la recta numérica en tres intervalos: 1. \( (-\infty, 1) \) 2. \( (1, 6) \) 3. \( (6, +\infty) \) ### 3. Analizar el signo de la expresión en cada intervalo Seleccionamos un número de prueba en cada intervalo para determinar el signo de la expresión \( x^{2} - 7x + 6 \). - **Intervalo \( (-\infty, 1) \):** Tomemos \( x = 0 \): \[ 0^{2} - 7(0) + 6 = 6 > 0 \] - **Intervalo \( (1, 6) \):** Tomemos \( x = 3 \): \[ 3^{2} - 7(3) + 6 = 9 - 21 + 6 = -6 < 0 \] - **Intervalo \( (6, +\infty) \):** Tomemos \( x = 7 \): \[ 7^{2} - 7(7) + 6 = 49 - 49 + 6 = 6 > 0 \] ### 4. Determinar la solución de la desigualdad La desigualdad \( x^{2} - 7x + 6 \leq 0 \) se cumple donde la expresión es negativa o igual a cero. Según el análisis anterior, esto ocurre en el intervalo \( [1, 6] \). ### **Respuesta final:** \[ x \in [1, 6] \] Es decir, todos los valores de \( x \) entre 1 y 6, inclusive, satisfacen la desigualdad.

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Bonus Knowledge

To solve the inequality \( x^{2}-7 x+6 \leq 0 \), start by factoring it. The expression factors to \( (x - 1)(x - 6) \leq 0 \). This means that the critical points are \( x = 1 \) and \( x = 6 \). Testing intervals, we find the solution set includes values from 1 to 6, i.e., \( 1 \leq x \leq 6 \). Plotting the graph of the quadratic shows that it opens upwards, which helps visualize where it is less than or equal to zero. The curve crosses the x-axis at the critical points, indicating where the solution holds. To confirm, values within the intervals can be tested: picking \( x = 3 \) falls within the interval, and substituting gives a negative result, ensuring \( (x - 1)(x - 6) \leq 0 \).

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