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Q.
The value of a for which one root of the equation $x^2-(a+1) x+a^2+a-8=0$ exceeds 2 and other is less than 2 , are given by
Complex Numbers and Quadratic Equations
Solution:
As the roots are real and distinct
$ (a+1)^2-4\left(a^2+a-8\right)>0 $
$3 a^2+2 a-33< 0$
$\Rightarrow (3 a+11)(a-3)< 0 \Rightarrow-\frac{11}{3}< a< 3$(1)
Also, for
$\Rightarrow 2^2-2(a+1)+a^2+a-8< 0 $
$\Rightarrow a^2-a-6< 0 \Rightarrow(a-3)(a+2)< 0$
$\Rightarrow -2< a< 3$ (2)
From (1) and (2), we get $-2< a< 3$.
