For what value(s) of \( k \) will the equation \( -2 k=3 x^{2}+4 x \) have non-real roots? INSTRUCTION: Give your answer as an equation or inequality with a fraction (no decimal values). For example: \( k=4 \).

To determine the value(s) of \( k \) for which the equation \[ -2k = 3x^2 + 4x \] has non-real roots, we need to analyze the quadratic equation in the standard form \( ax^2 + bx + c = 0 \). Rearranging the equation gives us: \[ 3x^2 + 4x + 2k = 0 \] Here, we identify the coefficients: - \( a = 3 \) - \( b = 4 \) - \( c = 2k \) For a quadratic equation to have non-real roots, the discriminant must be less than zero. The discriminant \( D \) is given by: \[ D = b^2 - 4ac \] Substituting the values of \( a \), \( b \), and \( c \): \[ D = 4^2 - 4 \cdot 3 \cdot (2k) \] Calculating this gives: \[ D = 16 - 24k \] To find the values of \( k \) for which the roots are non-real, we set the discriminant less than zero: \[ 16 - 24k < 0 \] Now, we will solve this inequality for \( k \): \[ 16 < 24k \] Dividing both sides by 24: \[ \frac{16}{24} < k \] Simplifying the fraction: \[ \frac{2}{3} < k \] Thus, the solution can be expressed as: \[ k > \frac{2}{3} \] This means that for any value of \( k \) greater than \( \frac{2}{3} \), the equation will have non-real roots.