Question

The values of $p$ for which one root of the equation $x^2-(p+1) x+p^2+p=8$ exceeds 2 and the other is lesser than 2 are given by

• A. $3< $ p $< 10$
• B. p $\leq-2$
• C. $p \geq 10$
• D. $-2< $ p $< 3$

Answer 1

Answer: (D)

$ x ^2-( p +1) x + p ^2+ p -8=0$
$ f (2)<0$
$\Rightarrow 4-2( P +1)+ P ^2+ P -8<0 $
$\Rightarrow P ^2- P -6<0 \Rightarrow( P -3)( P -2)<0$
$\Rightarrow P \in(-2,-3)$