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\begin{tabular}{l} What are the roots of the equation? \\ \( \qquad-3 x+19=-x^{2}+1 \) \\ Answer \\ \( \begin{array}{l}\frac{3+9 i}{2} \\ \frac{-3+1 \sqrt{63}}{2} \\ \frac{3+i \sqrt{63}}{2}\end{array} \) \\ \hline Answer \end{tabular}

Ask by Wheeler Delgado. in the United States
Mar 10,2025

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The roots of the equation are \( x = \frac{3 + i\sqrt{63}}{2} \) and \( x = \frac{3 - i\sqrt{63}}{2} \).

Solución

We start with the equation \[ -3x+19=-x^2+1. \] **Step 1. Rearrange the equation** Add \(x^2\) to both sides and subtract \(1\) from both sides: \[ x^2-3x+19-1=0, \] which simplifies to \[ x^2-3x+18=0. \] **Step 2. Compute the discriminant** For a quadratic equation \(ax^2+bx+c=0\), the discriminant is \[ \Delta=b^2-4ac. \] Here, \(a=1\), \(b=-3\), and \(c=18\). Thus, \[ \Delta=(-3)^2-4\cdot1\cdot18=9-72=-63. \] **Step 3. Find the roots using the quadratic formula** The quadratic formula is \[ x=\frac{-b\pm\sqrt{\Delta}}{2a}. \] Substitute the values: \[ x=\frac{-(-3)\pm\sqrt{-63}}{2\cdot1}=\frac{3\pm\sqrt{-63}}{2}. \] Since \(\sqrt{-63}=i\sqrt{63}\), we have: \[ x=\frac{3\pm i\sqrt{63}}{2}. \] This can also be written as: \[ x=\frac{3\pm 3i\sqrt{7}}{2}. \] **Final Answer** \[ x=\frac{3 \pm i\sqrt{63}}{2}\quad\text{or}\quad x=\frac{3 \pm 3i\sqrt{7}}{2}. \]

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The Deep Dive

To tackle this equation, let's first rearrange it into standard form. Rewriting \(-3x + 19 = -x^2 + 1\) gives us the quadratic equation \(x^2 - 3x + 18 = 0\). Now, applying the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = -3\), and \(c = 18\) leads to complex solutions, showcasing the beauty of mathematics where even "imaginary" numbers lend clarity and provide insight into the nature of the roots! Understanding the roots helps not just in solving equations but also in graphing them! This quadratic equation describes a parabola that opens upward and has complex roots, which means its vertex is above the x-axis, reflecting that it doesn't intersect with the line \(y=0\). This is a common scenario in algebra where visualizing the graph can aid in comprehending the behavior of functions, especially when you encounter complex or imaginary numbers!

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