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Question 25 Solve equation by using the quadratic formula. List the solutions, separated by commas. Enter exact solutions. \( 3 k^{2}+k-8=0 \) \( k=\square \) Question Help: Video D Post to forum Submit Question

Ask by Weston Mcguire. in the United States
Mar 09,2025

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Answer

\( k = \frac{-1 + \sqrt{97}}{6}, \frac{-1 - \sqrt{97}}{6} \)

Solution

We start with the quadratic equation: \[ 3k^2 + k - 8 = 0 \] **Step 1.** Identify the coefficients \(a\), \(b\), and \(c\): \[ a = 3,\quad b = 1,\quad c = -8 \] **Step 2.** Write the quadratic formula: \[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] **Step 3.** Substitute the coefficients: \[ k = \frac{-1 \pm \sqrt{1^2 - 4(3)(-8)}}{2(3)} \] **Step 4.** Simplify the expression inside the square root (the discriminant): \[ 1^2 - 4(3)(-8) = 1 + 96 = 97 \] **Step 5.** Write the final solutions: \[ k = \frac{-1 \pm \sqrt{97}}{6} \] Thus, the two exact solutions are: \[ k = \frac{-1 + \sqrt{97}}{6},\quad \frac{-1 - \sqrt{97}}{6} \]

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To solve the quadratic equation \(3k^{2} + k - 8 = 0\) using the quadratic formula, we'll use the formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), and \(c = -8\). First, calculate the discriminant: \(b^2 - 4ac = 1^2 - 4(3)(-8) = 1 + 96 = 97\). Now, substituting into the formula gives: \[ k = \frac{-1 \pm \sqrt{97}}{2 \cdot 3} = \frac{-1 \pm \sqrt{97}}{6} \] The exact solutions are \(k = \frac{-1 + \sqrt{97}}{6}, \frac{-1 - \sqrt{97}}{6}\). Did you know that quadratic equations have been around since ancient civilizations? Babylonians, as early as 2000 BC, solved quadratic equations, although their methods weren’t as formalized as we see today. They often solved practical problems like land area calculations, laying a foundation for algebra! In the real world, quadratic equations frequently pop up! From physics—like calculating projectile motion—to finance—evaluating profit scenarios—this equation type can help us model and solve various complex situations. They provide critical insight into maximizing profits or predicting the trajectory of objects, proving that math isn’t just for classrooms!

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