Question Number 136124 by bramlexs22 last updated on 18/Mar/21
$$ \\ $$Given a quadratic function f(x) =3-4k-(k+3) x-x^2, where k is a constant, is always negative when p<k<q. What is the value of p and q?
Answered by EDWIN88 last updated on 18/Mar/21
$$\mathrm{f}\left(\mathrm{x}\right)=−\mathrm{x}^{\mathrm{2}} −\left(\mathrm{k}+\mathrm{3}\right)\mathrm{x}+\mathrm{3}−\mathrm{4k}\:<\:\mathrm{0}\:\forall\mathrm{x}\in\mathbb{R} \\ $$$$\mathrm{taking}\:\mathrm{discriminant}\:\Delta<\mathrm{0} \\ $$$$\therefore\:\left(\mathrm{k}+\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{1}\right)\left(\mathrm{3}−\mathrm{4k}\right)<\mathrm{0} \\ $$$$\mathrm{k}^{\mathrm{2}} +\mathrm{6k}+\mathrm{9}+\mathrm{12}−\mathrm{16k}<\mathrm{0} \\ $$$$\mathrm{k}^{\mathrm{2}} −\mathrm{10k}+\mathrm{21}<\mathrm{0}\:\Rightarrow\left(\mathrm{k}−\mathrm{7}\right)\left(\mathrm{k}−\mathrm{3}\right)<\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{3}\:<\:\mathrm{k}\:<\:\mathrm{7}\:\mathrm{so}\:\begin{cases}{\mathrm{p}=\mathrm{3}}\\{\mathrm{q}=\mathrm{7}}\end{cases} \\ $$