Evaluate \( \frac{x}{y} \) if \( 2 x^{2}-x y-3 y^{2}=0 \)

To evaluate \( \frac{x}{y} \), we start with the given equation: \[ 2x^2 - xy - 3y^2 = 0. \] Let's make the substitution \( r = \frac{x}{y} \), which leads us to express \( x = ry \). Substituting \( x \) in terms of \( r \) and \( y \) into the original equation gives us: \[ 2(ry)^2 - (ry)y - 3y^2 = 0. \] Simplifying this, we have: \[ 2r^2y^2 - ry^2 - 3y^2 = 0. \] Factoring out \( y^2 \) (assuming \( y \neq 0 \)), we arrive at: \[ y^2 (2r^2 - r - 3) = 0. \] Since \( y^2 \neq 0 \), we need to solve the quadratic equation: \[ 2r^2 - r - 3 = 0. \] Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 2 \), \( b = -1 \), and \( c = -3 \): \[ r = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 24}}{4} = \frac{1 \pm \sqrt{25}}{4} = \frac{1 \pm 5}{4}. \] This gives us the two potential solutions for \( r \): 1. \( r = \frac{6}{4} = \frac{3}{2} \) 2. \( r = \frac{-4}{4} = -1 \) Thus, the possible values for \( \frac{x}{y} \) are \( \frac{3}{2} \) and \( -1 \).