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  4. If x2+(a-b) x+(1-a-b)=0, a, b ∈ R then the value of 'a' for which both roots of the equation are real and unequal ∀ b ∈ R is

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Q. If $x^2+(a-b) x+(1-a-b)=0, a, b \in R$ then the value of 'a' for which both roots of the equation are real and unequal $\forall b \in R$ is

Complex Numbers and Quadratic Equations

Solution:

$x^2+(a-b) x+(1-a-b)=0$
$\because D >0$
$\Rightarrow (a-b)^2-4 \times 1 \times(1-a-b)>0$
$a^2+b^2-2 a b-4+4 a+4 b>0$
$\because b^2+2 b(2-a)+\left(a^2+4 a-4\right)>0$
$\because 4(2-a)^2-4 \times 1 \times\left(a^2+4 a-4\right)<0$
$4+a^2-4 a-a^2-4 a+4<0$
$8 a-8 >0$
$a >1$