Question

Consider the equation given below.$x^2+(a-b)x+(1-a-b)=0,b\in R$.Find the condition on ${}^{\prime }{a}^{\prime }$ for which both roots of the equation are real and unequal $\forall b\in R$.
Roots are imaginary $\forall bϵR$

Answer 1

Consider the given equation.

$x^2+(a-b)x+(1-a-b)=0$

If both the roots are to be real and unequal. we know that

$b^2-4ac>0$

$(a-b{)}^2-4\times (1-a-b)>0$

$a^2+b^2-2ab-4+4a+4b>0$

$b^2-(2a-4)b+a^2+4a-4>0$

Since, the above equation is always greater than $0$, it will not have any real roots. Thus, we have

$(2a-4{)}^2-4(a^2+4a-4)<0$

$4a^2-16a+16-4a^2-16a+16<0$

$-32a<-32$

$a<1$

Thus, if the equation has to always have real and unequal roots $\forall b\in R$, $a<1$.