(5 pts) Analyze the studentrs work in solving the system of equations below. (i) Describe their error an (ii) solve the problem correctly. \[ \begin{array}{l} y=2 x^{2}+3 \\ y=3 x+1 \end{array} \] \[ \begin{aligned} y & =2 x^{2}+3 \\ -(y & =3 x+1) \\ 0 & =-x^{2}+2 \\ x^{2} & =2 \\ x & = \pm \sqrt{2} \approx \pm 1.41 \end{aligned} \]

The student's error lies in their approach to eliminating \( y \). They manipulated the equations correctly but didn't account for the fact that one equation is quadratic and the other is linear, leading to the misunderstanding of integrating both equations. Instead of directly substituting or correctly rearranging the first equation and setting it equal to the second one, they incorrectly concluded that \( 0 = -x^{2} + 2 \). To solve the system correctly, let's substitute the expression for \( y \) from the second equation into the first: \[ 3x + 1 = 2x^2 + 3. \] Now, rearranging terms gives: \[ 2x^2 - 3x + 2 = 0. \] Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we substitute \( a = 2, b = -3, c = 2 \): \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(2)(2)}}{2(2)} = \frac{3 \pm \sqrt{9 - 16}}{4} = \frac{3 \pm \sqrt{-7}}{4}. \] Since the discriminant is negative, there are no real solutions to this system, indicating that the curves do not intersect.