19. Show that the equation $x^{2}-(2 a+b) x+a b=0$ has real roots for all values of $a$ and $b$

Step 1 ：Given the equation \(x^{2}-(2 a+b) x+a b=0\), we need to show that it has real roots for all values of \(a\) and \(b\).

Step 2 ：To do this, we will prove that the discriminant of the quadratic equation is greater than or equal to 0. The discriminant is given by the formula: \(D = B^2 - 4AC\), where \(A\), \(B\), and \(C\) are the coefficients of the quadratic equation \(Ax^2 + Bx + C = 0\). In our case, \(A = 1\), \(B = -(2a + b)\), and \(C = ab\).

Step 3 ：Let's calculate the discriminant and check if it's greater than or equal to 0: \(D = (-2a - b)^2 - 4(1)(ab)\)

Step 4 ：Simplifying the discriminant, we get: \(D = 4a^2 + b^2\)

Step 5 ：\(\boxed{D = 4a^2 + b^2}\), which is always greater than or equal to 0 for all values of \(a\) and \(b\), since the square of any real number is non-negative.

Step 6 ：Therefore, the equation \(x^2 - (2a + b)x + ab = 0\) has real roots for all values of \(a\) and \(b\).