Prove that the roots of the equation \({x}^{2}-\left(a+b\right)x+ab-{p}^{2}=0\) are real for all real values of \(a\), \(b\) and \(p\).

We need to prove that \(\Delta \geq 0\).

\begin{align*}
\Delta
&=(-a-b)^2 - 4(ab-p^2)\\
&=a^2 + 2ab + b^2 - 4ab + 4p^2\\
&=a^2 - 2ab + b^2 + 4p^2\\
&=(a-b)^2 + 4p^2
\end{align*}

\(\Delta \geq 0\) for all real values of \(a\), \(b\) and \(p\). Therefore the roots are real for all real values of \(a\), \(b\) and \(p\).

When will the roots of the equation be equal?

The roots are equal when \(\Delta = 0\), that is when \(a=b\) and \(p=0\).

If \(b\) and \(c\) can take on only the values \(\text{1}\), \(\text{2}\) or \(\text{3}\), determine all pairs (\(b; c\)) such that \({x}^{2}+bx+c=0\) has real roots.

[IEB, Nov. 2005, HG]

We need to find the values of \(a\), \(b\) and \(c\) for which \(\Delta \geq 0\).

\begin{align*}
a&=1\\
b&=\text{1,2} \text{ or }3\\
c&=\text{1,2} \text{ or }3\\
\\
\Delta &= b^2-4ac\\
&= b^2 - 4(1)c
\end{align*}

Possible pair values of \((b;c)\): \((1;1),~(1;2),~(1;3),~(2;1),~(2;2),~(2;3),~(3;1),~(3;2),~(3;3)\).
Corresponding values of \(\Delta\): \((\Delta < 0),~(\Delta < 0),~(\Delta < 0),~(\Delta = 0),~(\Delta < 0),~(\Delta < 0),~(\Delta >0),~(\Delta > 0),~(\Delta < 0)\)

\(\Delta ≥ 0\) (and therefore the roots are real) for \((b;c) = (2;1),~(3;1),~(3;2)\)