Question

If the roots of the equation $x^2-(p+1)x+p^2+p-8=0$ are on opposite sides with respect to the line x = 2 then the range of ‘p’ is

• A. $(-\frac{11}{3},3)$
• B. $(-2,3)$
• C. $(-\infty ,-2)\cup (3,\infty )$
• D. $\varphi$

Answer 1

The correct option is B $(-2,3)$

Here D > 0 ; $\alpha$ < 2 < $\beta$
$D=(p+1{)}^2-4(p^2+p-8)$ > $0$
$\Rightarrow 3p^2+2p-33$ < $0$
$\Rightarrow (3p+11)(p-3)$ < $0$ $\Rightarrow pϵ(-\frac{11}{3},3)....\left(1\right)$
$\Rightarrow f\left(2\right)$ < $0$
Also $f\left(2\right)=4-2(p+1)+p^2+p-8$ < $0....\left(2\right)$
Common solution of (1) and (2) is (-2,3).