Show that the roots of the equation \( x^{2}+(1-a) x=a \) are real for all real vac \[ x^{2}+(1-a) x=a \]

To show that the roots of the equation \( x^{2} + (1-a)x = a \) are real for all real values of \( a \), we need to analyze the discriminant of the quadratic equation in the standard form \( Ax^2 + Bx + C = 0 \). First, we can rearrange the equation to standard quadratic form: \[ x^2 + (1-a)x - a = 0 \] Here, we identify: - \( A = 1 \) - \( B = 1 - a \) - \( C = -a \) The discriminant \( D \) of a quadratic equation \( Ax^2 + Bx + C = 0 \) is given by: \[ D = B^2 - 4AC \] Substituting in our values for \( A \), \( B \), and \( C \): \[ D = (1-a)^2 - 4(1)(-a) \] Calculating gives us: \[ D = (1 - a)^2 + 4a \] Now expand \( (1 - a)^2 \): \[ (1 - a)^2 = 1 - 2a + a^2 \] So, \[ D = 1 - 2a + a^2 + 4a = a^2 + 2a + 1 \] Notice that \( a^2 + 2a + 1 \) can be factored as: \[ D = (a + 1)^2 \] Now, since the square of any real number is non-negative, we have: \[ (a + 1)^2 \geq 0 \] This means \( D \) is always greater than or equal to zero for all real values of \( a \). Since the discriminant \( D \) is non-negative, it follows that the quadratic equation \( x^{2} + (1-a)x - a = 0 \) has real roots for all real values of \( a \). Thus, we have shown that the roots of the equation are indeed real for all real values of \( a \).