EXERCISE 8 1. Without solving each of the following equations, discuss the nature of the roots: \( \begin{array}{lll}\text { (a) } x^{2}+x+1=0 & \text { (b) } x^{2}=2(x+1) & \text { (c) } x^{2}=4 x \\ \text { (d) }-2 x^{2}-16 x-32=0 & \text { (e) }(x-2)(2 x-1)=5 & \text { (f) } 3 x+7=\frac{5}{x}\end{array} \) 2. Show that the roots of the equation \( m x(x-4)=-4 m \) are equal for all real values of \( m \). Show that the roots of the equation \( x(x-3 m)=-5 m^{2} \) are non-real if \( m \neq 0 \)

To assess the nature of the roots without directly solving the equations, we can analyze the discriminant (in the quadratic cases) or specific properties of each equation. For the first part: (a) The roots of \( x^{2}+x+1=0 \) are non-real because the discriminant, \( D = b^2 - 4ac = 1 - 4 \), is negative. (b) Rearranging gives \( x^2 - 2x - 2 = 0 \), whose discriminant \( D = (-2)^2 - 4(1)(-2) = 4 + 8 = 12 \), indicates two distinct real roots. (c) The rearrangement leads to \( x^2 - 4x = 0 \), and factoring gives \( x(x-4)=0 \), revealing roots \( x=0 \) and \( x=4 \) (two distinct real roots). (d) The equation can be simplified to \( x^{2} + 8 x + 16 = 0 \). The discriminant here is \( D = 64 - 64 = 0 \), signifying one real double root. (e) Expanding gives \( 2x^2 - 5x + 4 = 0 \), and its discriminant \( D = 25 - 32 = -7 \) reveals non-real roots. (f) Rearranging yields \( 3x^2 + 7x - 5 = 0 \), where \( D = 49 + 60 = 109 \), indicating two distinct real roots. For the second part: The equation \( m x(x-4) = -4m \) simplifies to \( x^2 - 4x + 4 = 0 \), representing \( (x-2)^2 = 0 \) which shows that it yields a double root \( x = 2 \) for any real value of \( m \) (as long as \( m \neq 0 \)). In contrast, the equation \( x(x - 3m) = -5m^2 \) can be rearranged to \( x^2 - 3mx + 5m^2 = 0 \). The discriminant in this case \( D = (-3m)^2 - 4(1)(5m^2) = 9m^2 - 20m^2 = -11m^2 \) is negative for all \( m \neq 0 \), implying that the roots are indeed non-real.