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Q.
If exactly one root of equation $x^2-(p+1) x-p^2=0$ lie between 1 and 4 then number of integral values of $p$ is -
Complex Numbers and Quadratic Equations
Solution:
$f(1) f(4)< 0$
$\left(-p-p^2\right)\left(12-4 p-p^2\right)< 0$
$p(p+1)(p+6)(p-2)< 0$
so $p \in(-6,-1) \cup(0,2)$
number of integral values of $p$ is $5 $
