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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 Lynch Patel. in El Salvador
Jan 24,2025

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Answer

**Soluciones de las Ecuaciones Cuarticas:** - **a)** \( x = -2, -1, 1, 2 \) - **b)** \( x = -3, -2, 2, 3 \) - **c)** \( x = -5, -2, 2, 5 \) - **d)** - Reales: \( x = -3, 3 \) - Complejas: \( x = -i, i \) - **e)** \( x = -2i, -i, i, 2i \) - **f)** \( x = -\sqrt{3}i, -i, i, \sqrt{3}i \)

Solution

Claro, resolveremos cada una de las ecuaciones cuarticas dadas paso a paso. Utilizaremos la sustitución \( y = x^2 \) para simplificar las ecuaciones a cuadráticas y luego resolveremos para \( y \) y posteriormente para \( x \). ### a) \( x^{4} - 5x^{2} + 4 = 0 \) **Paso 1: Sustitución** Sea \( y = x^{2} \). Entonces, la ecuación se convierte en: \[ y^{2} - 5y + 4 = 0 \] **Paso 2: Resolver la ecuación cuadrática** \[ y^{2} - 5y + 4 = 0 \] Usando la fórmula cuadrática: \[ y = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2} \] Entonces: \[ y_1 = \frac{5 + 3}{2} = 4 \] \[ y_2 = \frac{5 - 3}{2} = 1 \] **Paso 3: Resolver para \( x \)** Para \( y = 4 \): \[ x^{2} = 4 \Rightarrow x = \pm 2 \] Para \( y = 1 \): \[ x^{2} = 1 \Rightarrow x = \pm 1 \] **Soluciones:** \[ x = -2, -1, 1, 2 \] --- ### b) \( x^{4} - 13x^{2} + 36 = 0 \) **Paso 1: Sustitución** Sea \( y = x^{2} \). Entonces: \[ y^{2} - 13y + 36 = 0 \] **Paso 2: Resolver la ecuación cuadrática** \[ y = \frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2} \] Entonces: \[ y_1 = \frac{13 + 5}{2} = 9 \] \[ y_2 = \frac{13 - 5}{2} = 4 \] **Paso 3: Resolver para \( x \)** Para \( y = 9 \): \[ x^{2} = 9 \Rightarrow x = \pm 3 \] Para \( y = 4 \): \[ x^{2} = 4 \Rightarrow x = \pm 2 \] **Soluciones:** \[ x = -3, -2, 2, 3 \] --- ### c) \( x^{4} - 29x^{2} + 100 = 0 \) **Paso 1: Sustitución** Sea \( y = x^{2} \). Entonces: \[ y^{2} - 29y + 100 = 0 \] **Paso 2: Resolver la ecuación cuadrática** \[ y = \frac{29 \pm \sqrt{841 - 400}}{2} = \frac{29 \pm \sqrt{441}}{2} = \frac{29 \pm 21}{2} \] Entonces: \[ y_1 = \frac{29 + 21}{2} = 25 \] \[ y_2 = \frac{29 - 21}{2} = 4 \] **Paso 3: Resolver para \( x \)** Para \( y = 25 \): \[ x^{2} = 25 \Rightarrow x = \pm 5 \] Para \( y = 4 \): \[ x^{2} = 4 \Rightarrow x = \pm 2 \] **Soluciones:** \[ x = -5, -2, 2, 5 \] --- ### d) \( x^{4} - 8x^{2} - 9 = 0 \) **Paso 1: Sustitución** Sea \( y = x^{2} \). Entonces: \[ y^{2} - 8y - 9 = 0 \] **Paso 2: Resolver la ecuación cuadrática** \[ y = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2} \] Entonces: \[ y_1 = \frac{8 + 10}{2} = 9 \] \[ y_2 = \frac{8 - 10}{2} = -1 \] **Paso 3: Resolver para \( x \)** Para \( y = 9 \): \[ x^{2} = 9 \Rightarrow x = \pm 3 \] Para \( y = -1 \): \[ x^{2} = -1 \Rightarrow x = \pm i \] **Soluciones:** - **Reales:** \( x = -3, 3 \) - **Complejas:** \( x = -i, i \) --- ### e) \( x^{4} + 5x^{2} + 4 = 0 \) **Paso 1: Sustitución** Sea \( y = x^{2} \). Entonces: \[ y^{2} + 5y + 4 = 0 \] **Paso 2: Resolver la ecuación cuadrática** \[ y = \frac{-5 \pm \sqrt{25 - 16}}{2} = \frac{-5 \pm 3}{2} \] Entonces: \[ y_1 = \frac{-5 + 3}{2} = -1 \] \[ y_2 = \frac{-5 - 3}{2} = -4 \] **Paso 3: Resolver para \( x \)** Para \( y = -1 \): \[ x^{2} = -1 \Rightarrow x = \pm i \] Para \( y = -4 \): \[ x^{2} = -4 \Rightarrow x = \pm 2i \] **Soluciones:** \[ x = -2i, -i, i, 2i \] --- ### f) \( x^{4} + 4x^{2} + 3 = 0 \) **Paso 1: Sustitución** Sea \( y = x^{2} \). Entonces: \[ y^{2} + 4y + 3 = 0 \] **Paso 2: Resolver la ecuación cuadrática** \[ y = \frac{-4 \pm \sqrt{16 - 12}}{2} = \frac{-4 \pm 2}{2} \] Entonces: \[ y_1 = \frac{-4 + 2}{2} = -1 \] \[ y_2 = \frac{-4 - 2}{2} = -3 \] **Paso 3: Resolver para \( x \)** Para \( y = -1 \): \[ x^{2} = -1 \Rightarrow x = \pm i \] Para \( y = -3 \): \[ x^{2} = -3 \Rightarrow x = \pm \sqrt{3}i \] **Soluciones:** \[ x = -\sqrt{3}i, -i, i, \sqrt{3}i \] --- **Resumen de Soluciones:** - **a)** \( x = -2, -1, 1, 2 \) - **b)** \( x = -3, -2, 2, 3 \) - **c)** \( x = -5, -2, 2, 5 \) - **d)** - Reales: \( x = -3, 3 \) - Complejas: \( x = -i, i \) - **e)** \( x = -2i, -i, i, 2i \) - **f)** \( x = -\sqrt{3}i, -i, i, \sqrt{3}i \)

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To solve these quartic equations, it's often helpful to use a substitution method. By letting \( y = x^2 \), we can transform the equations into quadratic forms. For instance, the first equation \( x^{4}-5x^{2}+4=0 \) becomes \( y^2 - 5y + 4 = 0 \). This can be solved using the quadratic formula or factoring to find the values of \( y \). Now, when you have the values of \( y \), simply revert to \( x \) by taking the square roots. Remember to consider both the positive and negative roots, as both are valid solutions! This method continues to apply for the other equations by similarly substituting \( y \) for each \( x^{2} \). Keep it fun and remember to check for any extraneous solutions after substituting back!

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