QUESTION 3 \( 3.1 \quad \) For which value(s) of \( \boldsymbol{m} \) will the equation \( 2 x(x+1)+m=x \) have non-real roots?

To determine the value(s) of \( m \) for which the equation \( 2 x(x+1) + m = x \) has non-real roots, we first rearrange the equation into standard quadratic form: \[ 2x^2 + 2x + m - x = 0 \] \[ 2x^2 + x + m = 0 \] Next, we will apply the discriminant condition. For a quadratic equation \( ax^2 + bx + c = 0 \), the discriminant \( D = b^2 - 4ac \) indicates the nature of the roots: - If \( D > 0 \), the equation has two distinct real roots. - If \( D = 0 \), there is exactly one real root (a repeated root). - If \( D < 0 \), there are non-real (complex) roots. For our equation, \( a = 2 \), \( b = 1 \), and \( c = m \). Thus, the discriminant \( D \) is: \[ D = (1)^2 - 4(2)(m) = 1 - 8m \] We want the discriminant to be less than zero for non-real roots: \[ 1 - 8m < 0 \] Solving for \( m \): \[ -8m < -1 \] \[ m > \frac{1}{8} \] Therefore, the equation \( 2x(x+1) + m = x \) will have non-real roots for all values of \( m \) greater than \( \frac{1}{8} \).