What are the two solutions of \( 2 x^{2}=-x^{2}-5 x-1 \) ? the \( y \)-coordinates of the intersection points of the graphs of \( y=2 x^{2} \) and \( y=-x^{2}-5 x-1 \) the \( x \)-coordinates of the intersection points of the graphs of \( y=2 x^{2} \) and \( y=-x^{2}-5 x-1 \) and

Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}=-x^{2}-5x-1\) - step1: Move the expression to the left side: \(3x^{2}+5x+1=0\) - step2: Solve using the quadratic formula: \(x=\frac{-5\pm \sqrt{5^{2}-4\times 3}}{2\times 3}\) - step3: Simplify the expression: \(x=\frac{-5\pm \sqrt{5^{2}-4\times 3}}{6}\) - step4: Simplify the expression: \(x=\frac{-5\pm \sqrt{13}}{6}\) - step5: Separate into possible cases: \(\begin{align}&x=\frac{-5+\sqrt{13}}{6}\\&x=\frac{-5-\sqrt{13}}{6}\end{align}\) - step6: Rewrite the fraction: \(\begin{align}&x=\frac{-5+\sqrt{13}}{6}\\&x=-\frac{5+\sqrt{13}}{6}\end{align}\) - step7: Rewrite: \(x_{1}=-\frac{5+\sqrt{13}}{6},x_{2}=\frac{-5+\sqrt{13}}{6}\) To find the solutions of the equation \( 2x^{2} = -x^{2} - 5x - 1 \), we first rearrange it into standard form: \[ 2x^{2} + x^{2} + 5x + 1 = 0 \] This simplifies to: \[ 3x^{2} + 5x + 1 = 0 \] Now, we can solve this quadratic equation. The solutions are: \[ x_{1} = -\frac{5 + \sqrt{13}}{6}, \quad x_{2} = -\frac{5 - \sqrt{13}}{6} \] ### \( y \)-coordinates of the intersection points To find the \( y \)-coordinates of the intersection points, we substitute the \( x \)-coordinates back into either of the original equations. We'll use \( y = 2x^{2} \). 1. For \( x_{1} = -\frac{5 + \sqrt{13}}{6} \): \[ y_{1} = 2\left(-\frac{5 + \sqrt{13}}{6}\right)^{2} \] 2. For \( x_{2} = -\frac{5 - \sqrt{13}}{6} \): \[ y_{2} = 2\left(-\frac{5 - \sqrt{13}}{6}\right)^{2} \] Now, let's calculate these \( y \)-coordinates. ### Calculating \( y_{1} \) and \( y_{2} \) 1. For \( y_{1} \): \[ y_{1} = 2\left(-\frac{5 + \sqrt{13}}{6}\right)^{2} = 2 \cdot \frac{(5 + \sqrt{13})^{2}}{36} = \frac{2(25 + 10\sqrt{13} + 13)}{36} = \frac{2(38 + 10\sqrt{13})}{36} = \frac{19 + 5\sqrt{13}}{18} \] 2. For \( y_{2} \): \[ y_{2} = 2\left(-\frac{5 - \sqrt{13}}{6}\right)^{2} = 2 \cdot \frac{(5 - \sqrt{13})^{2}}{36} = \frac{2(25 - 10\sqrt{13} + 13)}{36} = \frac{2(38 - 10\sqrt{13})}{36} = \frac{19 - 5\sqrt{13}}{18} \] ### Summary of Results - The \( x \)-coordinates of the intersection points are: - \( x_{1} = -\frac{5 + \sqrt{13}}{6} \) - \( x_{2} = -\frac{5 - \sqrt{13}}{6} \) - The \( y \)-coordinates of the intersection points are: - \( y_{1} = \frac{19 + 5\sqrt{13}}{18} \) - \( y_{2} = \frac{19 - 5\sqrt{13}}{18} \)