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Solve for \( z \). \( 4 z^{2}-7 z-5=0 \) \( z=\frac{7+\sqrt{129}}{8} \) or \( z=\frac{7-\sqrt{129}}{8} \) \( z=\frac{-7+\sqrt{129}}{8} \) or \( z=\frac{-7-\sqrt{129}}{8} \) \( z=\frac{7+\sqrt{69}}{8} \) or \( z=\frac{7-\sqrt{69}}{8} \) \( z=\frac{-7+\sqrt{69}}{8} \) or \( z=\frac{-7-\sqrt{69}}{8} \)

Ask by Barber Wang. in the United States
Jan 24,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solutions are \( z=\frac{7-\sqrt{129}}{8} \) and \( z=\frac{7+\sqrt{129}}{8} \).

Solution

Solve the equation \( 4z^{2}-7z-5=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(4z^{2}-7z-5=0\) - step1: Solve using the quadratic formula: \(z=\frac{7\pm \sqrt{\left(-7\right)^{2}-4\times 4\left(-5\right)}}{2\times 4}\) - step2: Simplify the expression: \(z=\frac{7\pm \sqrt{\left(-7\right)^{2}-4\times 4\left(-5\right)}}{8}\) - step3: Simplify the expression: \(z=\frac{7\pm \sqrt{129}}{8}\) - step4: Separate into possible cases: \(\begin{align}&z=\frac{7+\sqrt{129}}{8}\\&z=\frac{7-\sqrt{129}}{8}\end{align}\) - step5: Rewrite: \(z_{1}=\frac{7-\sqrt{129}}{8},z_{2}=\frac{7+\sqrt{129}}{8}\) The solutions to the equation \(4z^{2}-7z-5=0\) are \(z=\frac{7-\sqrt{129}}{8}\) and \(z=\frac{7+\sqrt{129}}{8}\).

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Beyond the Answer

To solve the quadratic equation \( 4z^2 - 7z - 5 = 0 \) using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we set \( a = 4 \), \( b = -7 \), and \( c = -5 \). Calculating the discriminant: \[ b^2 - 4ac = (-7)^2 - 4(4)(-5) = 49 + 80 = 129 \] Now substituting into the formula gives us: \[ z = \frac{7 \pm \sqrt{129}}{2 \cdot 4} = \frac{7 \pm \sqrt{129}}{8} \] Thus, the correct solutions are \( z = \frac{7 + \sqrt{129}}{8} \) or \( z = \frac{7 - \sqrt{129}}{8} \). Exploring the historical context, quadratic equations date back to ancient civilizations! The Babylonians (around 2000 BC) had methods for solving quadratic-type problems, but they didn’t necessarily use an algebraic framework. It wasn't until later that mathematicians like Al-Khwarizmi formalized and generalized these solutions, marking a significant milestone in the development of algebra. For real-world applications, solving quadratic equations is super useful! Whether you're calculating the trajectory of a soccer ball, optimizing area in landscaping, or figuring out profits in a business venture, quadratics pop up everywhere. Understanding how to solve them gives you a powerful tool for making informed decisions in various fields!

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