(d) For which value(s) of \( m \) will the following equations have non-real roots? \( \begin{array}{ll}\text { (1) } x^{2}-6 x=m & \text { (2) } x^{2}+m x+4=3 x \\ \text { (3) } \frac{x^{2}-5 x+5}{2 x^{2}-x+1}=\frac{1}{m} & \end{array} \)

**(1)** We start with \[ x^2 - 6x = m. \] Rewriting in standard form, we have \[ x^2 - 6x - m = 0. \] The discriminant is \[ \Delta = (-6)^2 - 4(1)(-m) = 36 + 4m. \] For the roots to be non-real, we need \(\Delta < 0\): \[ 36 + 4m < 0 \quad \Longrightarrow \quad 4m < -36 \quad \Longrightarrow \quad m < -9. \] --- **(2)** We start with \[ x^2 + m x + 4 = 3x. \] Rearrange the equation by subtracting \(3x\) from both sides: \[ x^2 + m x - 3x + 4 = 0 \quad \Longrightarrow \quad x^2 + (m-3)x + 4 = 0. \] The discriminant is \[ \Delta = (m-3)^2 - 4(1)(4) = (m-3)^2 - 16. \] For non-real roots, set \[ (m-3)^2 - 16 < 0 \quad \Longrightarrow \quad (m-3)^2 < 16. \] Taking square roots (and noting that this is equivalent to the absolute value inequality), \[ |m-3| < 4. \] This gives the inequality \[ -4 < m-3 < 4. \] Adding 3 to all parts, we obtain \[ -1 < m < 7. \] --- **(3)** We start with \[ \frac{x^2 - 5x + 5}{2x^2 - x + 1} = \frac{1}{m}. \] Since the denominator \(2x^2 - x + 1\) is always nonzero (its discriminant is negative), we can cross-multiply: \[ m(x^2 - 5x + 5) = 2x^2 - x + 1. \] Bring all terms to one side: \[ m x^2 - 5m x + 5m - 2x^2 + x - 1 = 0. \] Grouping like terms, \[ (m - 2)x^2 + (-5m + 1)x + (5m - 1) = 0. \] For this quadratic to have non-real roots, its discriminant must be negative. The discriminant is \[ \Delta = (-5m + 1)^2 - 4(m-2)(5m-1). \] First, expand \((-5m + 1)^2\): \[ (-5m + 1)^2 = 25m^2 - 10m + 1. \] Now expand the product: \[ 4(m-2)(5m-1) = 4\big(5m^2 - m - 10m + 2\big) = 4(5m^2 - 11m + 2) = 20m^2 - 44m + 8. \] So, \[ \Delta = 25m^2 - 10m + 1 - (20m^2 - 44m + 8) = 5m^2 + 34m - 7. \] Set the discriminant less than zero: \[ 5m^2 + 34m - 7 < 0. \] Solve the quadratic equation \[ 5m^2 + 34m - 7 = 0 \] using the quadratic formula: \[ m = \frac{-34 \pm \sqrt{34^2 - 4\cdot5\cdot(-7)}}{2\cdot5} = \frac{-34 \pm \sqrt{1156 + 140}}{10} = \frac{-34 \pm \sqrt{1296}}{10}. \] Since \(\sqrt{1296}=36\), we have \[ m = \frac{-34 \pm 36}{10}. \] This yields the two roots: \[ m = \frac{2}{10} = 0.2 \quad \text{and} \quad m = \frac{-70}{10} = -7. \] Since the coefficient of \(m^2\) is positive, the inequality \(5m^2 + 34m - 7 < 0\) holds between the roots. Therefore, \[ -7 < m < 0.2. \] --- **Final Answers:** - For equation (1): \( m < -9 \). - For equation (2): \( -1 < m < 7 \). - For equation (3): \( -7 < m < 0.2 \).