Question

If both the roots of the quadratic equation $x^2-2px+p^2+p-5=0$ are less than 3, then k lies in the interval

• A. $(-\infty ,1)$
• B. $(1,\infty )$
• C. $(1,3)$
• D. $(3,\infty )$

Answer 1

The correct option is A $(-\infty ,1)$

$\begin{array}{c}wehave\\ x^2-2px+ph2+p-5=0\\ xisrealthen,\\ D\ge 0\\ 4p^2-\mathrm{4.1.}\left(p^2+p-5\right)\ge 0\\ 4p^2-4p^2-4p+20\ge 0\\ -4p\ge -20\\ p\le 5----\left(1\right)\\ f\left(3\right)>0\left(bothrootsarelessthan3\right)\\ 3^2-3p.3+p^2+p-5>0\\ 9-6p+p^2+p-5>0\\ p^2-5p+4>0\\ \left(p-1\right)\left(p-4\right)>0\\ then\\ I\in \left(-\infty ,1\right)\cup \left(4,\infty \right)----\left(2\right)\\ also\\ \alpha +\beta <3\\ 2p<3\\ \therefore p<\frac{3}{2}---\left(3\right)\\ byeq{u}^n\left(1\right),\left(2\right),\&\left(3\right)weget\\ p\in \left(-\infty ,1\right).\\ Hence,optionAisthecorrectanswer.\end{array}$