If $x^{2}+(a-b) x+(1-a-b)=0$ where $a, b \in R$ then the minimum value of $a$ for which equation has

Question

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

Professor PC Thomas Classes · Answer

For unequal real roots $D>0$\ $\Rightarrow \quad(a-b)^{2}-4(1-a-b)>0$\ $\Rightarrow \quad b^{2}+b(4-2 a)+a^{2}+4 a-4>0$\ For the above quadratic expression to be true $\forall b \in R$\ Discriminant of its corresponding equation should be less than zero\ i.e., $(4-2 a)^{2}-4\left(a^{2}+4 a-4 ight)<0$\ $\Rightarrow \quad-32 a+32<0 \Rightarrow a>1$\