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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

f(x)=−x^2 −(k+3)x+3−4k < 0 ∀x∈R taking discriminant Δ<0 ∴ (k+3)^2 −4(−1)(3−4k)<0 k^2 +6k+9+12−16k<0 k^2 −10k+21<0 ⇒(k−7)(k−3)<0 we get 3 < k < 7 so { ((p=3)),((q=7)) :}

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

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