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\( \begin{array}{ll}x^{4}-5 x^{2}+4=0 & \text { b) } x^{4}-13 x^{2}+36= \\ x^{4}-29 x^{2}+100=0 & \text { d) } x^{4}-8 x^{2}-9=0 \\ x^{4}+5 x^{2}+4=0 & \text { f) } x^{4}+4 x^{2}+3=0\end{array} \)

Ask by Bradley Mccarthy. in El Salvador
Jan 24,2025

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**Soluciones:** - **b)** \( x = -3, -2, 2, 3 \) - **d)** \( x = -5, -2, 2, 5 \) - **f)** \( x = -i, i, -i\sqrt{3}, i\sqrt{3} \)

Solution

Claro, resolveremos las ecuaciones etiquetadas como **b)**, **d)** y **f)** paso a paso. --- ### **b) \( x^{4} - 13x^{2} + 36 = 0 \)** **Paso 1:** Realizamos un cambio de variable para simplificar la ecuación. Sea \( y = x^{2} \). Entonces la ecuación se convierte en: \[ y^{2} - 13y + 36 = 0 \] **Paso 2:** Factorizamos la ecuación cuadrática: \[ y^{2} - 13y + 36 = (y - 9)(y - 4) = 0 \] **Paso 3:** Encontramos los valores de \( y \): \[ y - 9 = 0 \quad \Rightarrow \quad y = 9 \] \[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \] **Paso 4:** Sustituimos de vuelta \( y = x^{2} \): \[ x^{2} = 9 \quad \Rightarrow \quad x = \pm 3 \] \[ x^{2} = 4 \quad \Rightarrow \quad x = \pm 2 \] **Soluciones:** \[ x = -3, \quad x = -2, \quad x = 2, \quad x = 3 \] --- ### **d) \( x^{4} - 29x^{2} + 100 = 0 \)** **Paso 1:** Realizamos un cambio de variable similar. Sea \( y = x^{2} \). La ecuación se convierte en: \[ y^{2} - 29y + 100 = 0 \] **Paso 2:** Factorizamos la ecuación cuadrática: \[ y^{2} - 29y + 100 = (y - 25)(y - 4) = 0 \] **Paso 3:** Encontramos los valores de \( y \): \[ y - 25 = 0 \quad \Rightarrow \quad y = 25 \] \[ y - 4 = 0 \quad \Rightarrow \quad y = 4 \] **Paso 4:** Sustituimos de vuelta \( y = x^{2} \): \[ x^{2} = 25 \quad \Rightarrow \quad x = \pm 5 \] \[ x^{2} = 4 \quad \Rightarrow \quad x = \pm 2 \] **Soluciones:** \[ x = -5, \quad x = -2, \quad x = 2, \quad x = 5 \] --- ### **f) \( x^{4} + 4x^{2} + 3 = 0 \)** **Paso 1:** Realizamos el cambio de variable. Sea \( y = x^{2} \). La ecuación se convierte en: \[ y^{2} + 4y + 3 = 0 \] **Paso 2:** Factorizamos la ecuación cuadrática: \[ y^{2} + 4y + 3 = (y + 1)(y + 3) = 0 \] **Paso 3:** Encontramos los valores de \( y \): \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \] \[ y + 3 = 0 \quad \Rightarrow \quad y = -3 \] **Paso 4:** Sustituimos de vuelta \( y = x^{2} \): \[ x^{2} = -1 \quad \Rightarrow \quad x = \pm i \] \[ x^{2} = -3 \quad \Rightarrow \quad x = \pm i\sqrt{3} \] **Soluciones:** \[ x = -i, \quad x = i, \quad x = -i\sqrt{3}, \quad x = i\sqrt{3} \] --- **Resumen de Soluciones:** - **b)** \( x = \pm 3, \quad x = \pm 2 \) - **d)** \( x = \pm 5, \quad x = \pm 2 \) - **f)** \( x = \pm i, \quad x = \pm i\sqrt{3} \)

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Bonus Knowledge

To solve these polynomial equations, a handy trick is to perform a substitution. Let's say \( y = x^2 \). This transforms the equations into quadratic form, making them easier to handle! For instance, the first equation \( x^{4} - 5x^{2} + 4 = 0 \) can be rewritten as \( y^{2} - 5y + 4 = 0 \). You can use the quadratic formula or factor it to find the roots. Keep in mind that solving for \( x \) will require taking the square roots of your \( y \) solutions, introducing both positive and negative options. Next, don’t forget to check under what conditions your original variable \( x \) reintegrates after deriving the \( y \) values from your equations. Any negative \( y \) values will not yield real solutions for \( x \)!

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