If $x^2+(a-b)x+(1-a-b)=0,$ where $a, b\in R,$ find the values of a for which the equation has unequal real roots for all values of b.

Since roots are real and unequal, so 

$\begin{array}{l} D>0\\ \Rightarrow (a-b{)}^2-4(1-a-b)>0\\ \Rightarrow (2-a{)}^2-\left(a^2+4a-4\right)<0\\ \Rightarrow a^2-4a+4-\left(a^2+4a-4\right)<0\end{array}$

Clearly, $D < 0$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(4-2a{)}^2-4\left(a^2+4a-4\right)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(2-a{)}^2-\left(a^2+4a-4\right)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{a}^2-4a+4-\left(a^2+4a-4\right)<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-4a+4-4a+4<0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-8a+8<0\\ \begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a-1>0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a>1\end{array}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a-1>0\end{array}$