Pregunta
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\begin{tabular}{|l|l} Solve. \\ \( 2 x^{2}-22=0 \)\end{tabular} \left\lvert\, \( \begin{array}{l}x=\square \\ \text { (Simplify your answer. Type an exact answer, using radicals as needed. } \\ \text { Express complex numbers in terms of } i \text {. Use a comma to separate answers } \\ \text { as needed.) }\end{array}\right. \)

Ask by Stephens Macdonald. in the United States
Feb 20,2025

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\( x = -\sqrt{11}, \sqrt{11} \)

Solución

Solve the quadratic equation by following steps: - step0: Solve using square roots: \(2x^{2}-22=0\) - step1: Add or subtract both sides: \(2x^{2}=22\) - step2: Divide both sides: \(\frac{2x^{2}}{2}=\frac{22}{2}\) - step3: Divide the numbers: \(x^{2}=11\) - step4: Simplify the expression: \(x=\pm \sqrt{11}\) - step5: Separate into possible cases: \(\begin{align}&x=\sqrt{11}\\&x=-\sqrt{11}\end{align}\) - step6: Rewrite: \(x_{1}=-\sqrt{11},x_{2}=\sqrt{11}\) To solve the equation \( 2x^{2} - 22 = 0 \), we can follow these steps: 1. **Isolate the quadratic term**: \[ 2x^{2} = 22 \] 2. **Divide both sides by 2**: \[ x^{2} = 11 \] 3. **Take the square root of both sides**: \[ x = \pm \sqrt{11} \] Thus, the solutions are: \[ x = -\sqrt{11}, \sqrt{11} \] So, the final answer is: \[ x = -\sqrt{11}, \sqrt{11} \]

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The Deep Dive

To solve the equation \( 2x^{2} - 22 = 0 \), start by adding 22 to both sides to get \( 2x^{2} = 22 \). Next, divide both sides by 2: \[ x^{2} = 11 \] Now, take the square root of both sides to find \( x \): \[ x = \pm \sqrt{11} \] So the solutions are: \[ x = \sqrt{11}, -\sqrt{11} \] ✨ Ever wondered about the history of the quadratic formula? It all dates back to ancient Babylonians around 2000 BC! They had methods to solve quadratic equations, albeit not in the general format we use today. They approximated the solutions for specific cases with remarkable accuracy, paving the way for future mathematicians. 🚀 In real-world applications, quadratic equations appear in various fields such as physics, finance, and engineering! For instance, if you're launching a projectile, its height can often be modeled by a quadratic equation. Finding the roots of that equation can help determine when the projectile hits the ground. How cool is that?

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