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What are the roots of the equation? \[ 3 x^{2}-16 x+15=3 \] Answer \( \begin{array}{l}\frac{2}{3} \text { and }-\frac{2}{3} \\ \frac{2}{3} \text { and }-6 \\ \begin{array}{l}\frac{16 \pm \sqrt{112}}{6} \\ \text { Answer }\end{array} \\ \frac{-16 \pm \sqrt{112}}{6}\end{array} \)

Ask by Bob Ortega. in the United States
Mar 10,2025

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Answer

The roots of the equation are \( x = \frac{8 - 2\sqrt{7}}{3} \) and \( x = \frac{8 + 2\sqrt{7}}{3} \).

Solution

Solve the equation \( 3x^{2}-16x+15=3 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(3x^{2}-16x+15=3\) - step1: Move the expression to the left side: \(3x^{2}-16x+12=0\) - step2: Solve using the quadratic formula: \(x=\frac{16\pm \sqrt{\left(-16\right)^{2}-4\times 3\times 12}}{2\times 3}\) - step3: Simplify the expression: \(x=\frac{16\pm \sqrt{\left(-16\right)^{2}-4\times 3\times 12}}{6}\) - step4: Simplify the expression: \(x=\frac{16\pm \sqrt{112}}{6}\) - step5: Simplify the expression: \(x=\frac{16\pm 4\sqrt{7}}{6}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{16+4\sqrt{7}}{6}\\&x=\frac{16-4\sqrt{7}}{6}\end{align}\) - step7: Simplify the expression: \(\begin{align}&x=\frac{8+2\sqrt{7}}{3}\\&x=\frac{16-4\sqrt{7}}{6}\end{align}\) - step8: Simplify the expression: \(\begin{align}&x=\frac{8+2\sqrt{7}}{3}\\&x=\frac{8-2\sqrt{7}}{3}\end{align}\) - step9: Rewrite: \(x_{1}=\frac{8-2\sqrt{7}}{3},x_{2}=\frac{8+2\sqrt{7}}{3}\) The roots of the equation \(3x^{2}-16x+15=3\) are \(x_{1}=\frac{8-2\sqrt{7}}{3}\) and \(x_{2}=\frac{8+2\sqrt{7}}{3}\).

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To find the roots of the equation \(3x^{2}-16x+15=3\), first, rearrange it to standard form: \(3x^{2}-16x+12=0\). Now, you can use the quadratic formula \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\) where \(a=3\), \(b=-16\), and \(c=12\). Plugging in these values, you get \(x=\frac{16 \pm \sqrt{(-16)^2 - 4(3)(12)}}{2(3)} = \frac{16 \pm \sqrt{112}}{6}\). If you simplify \(\sqrt{112}\) to \(4\sqrt{7}\) and continue, the roots can also be expressed as \(x=\frac{16 \pm 4\sqrt{7}}{6}=\frac{8 \pm 2\sqrt{7}}{3}\). Now you have the final answer: \(x = \frac{8 + 2\sqrt{7}}{3}\) and \(x = \frac{8 - 2\sqrt{7}}{3}\). In the realm of mathematics, quadratic equations have been pivotal since their inception! The Babylonians solved them as far back as 2000 BC using geometric methods. Fast forward to today, we still use those foundational concepts but express them in a more algebraic format, embracing everything from ancient tablets to computer algorithms that solve these equations in the blink of an eye. These equations aren’t just academic; they pop up in various real-world scenarios! From determining the flight path of a basketball to optimizing profits in business, quadratic equations are everywhere. Think of calculating the optimal dimensions for a garden or analyzing the trajectory of a roller coaster. Understanding how these equations work can help you make better decisions based on what's most efficient or effective in everyday life!

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