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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?\n

Answered by EDWIN88 last updated on 18/Mar/21

$$\mathrm{f}\left(\mathrm{x}\right)=-{\mathrm{x}}^2-(\mathrm{k}+3)\mathrm{x}+3-4\mathrm{k}<0\mathrm{\forall }\mathrm{x}\in \mathbb{R}$$ $$\mathrm{taking}\mathrm{discriminant}\mathrm{\Delta }<0$$ $$\therefore {(\mathrm{k}+3)}^2-4(-1)(3-4\mathrm{k})<0$$ $${\mathrm{k}}^2+6\mathrm{k}+9+12-16\mathrm{k}<0$$ $${\mathrm{k}}^2-10\mathrm{k}+21<0\Rightarrow (\mathrm{k}-7)(\mathrm{k}-3)<0$$ $$\mathrm{we}\mathrm{get}3<\mathrm{k}<7\mathrm{s}\mathrm{o}\{\begin{array}{l}\mathrm{p}=3\\ \mathrm{q}=7\end{array}$$

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