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

Step 1 ：Given the equation $x^2-(2a+b)x+ab=0$, 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 $A{x}^2+Bx+C=0$. 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 ：$\boxed{{\displaystyle D=4a^2+b^2}}$, 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 $b$.