Question Number 150006 by mathdanisur last updated on 08/Aug/21
$$\mathrm{If}\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{cannot}\:\mathrm{be}\:\:\mathrm{xy}.? \\ $$
Answered by dumitrel last updated on 08/Aug/21
![xy=k⇒x+(k/x)=(1/2)⇒2x^2 −x+2k=0⇒△≥0⇒ 1−16k≥0⇒k∈(−∞;(1/(16))]⇒k∉((1/(16));∞)](Q150015.png)
$${xy}={k}\Rightarrow{x}+\frac{{k}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{2}{k}=\mathrm{0}\Rightarrow\bigtriangleup\geqslant\mathrm{0}\Rightarrow \\ $$$$\mathrm{1}−\mathrm{16}{k}\geqslant\mathrm{0}\Rightarrow{k}\in\left(−\infty;\frac{\mathrm{1}}{\mathrm{16}}\right]\Rightarrow{k}\notin\left(\frac{\mathrm{1}}{\mathrm{16}};\infty\right) \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Thank}\:\mathrm{You}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\boldsymbol{\mathrm{Ser}},\:\mathrm{answer}:\:\mathrm{a}.\frac{\mathrm{1}}{\mathrm{15}}\:\:\:\mathrm{b}.\frac{\mathrm{1}}{\mathrm{16}}\:\:\:\mathrm{c}.\frac{\mathrm{1}}{\mathrm{17}}\:\:\mathrm{d}.\frac{\mathrm{1}}{\mathrm{18}}\:\:\mathrm{e}.\frac{\mathrm{1}}{\mathrm{19}} \\ $$
Commented by dumitrel last updated on 08/Aug/21
$${a}.\:\frac{\mathrm{1}}{\mathrm{15}} \\ $$
Commented by mathdanisur last updated on 08/Aug/21
$$\mathrm{Cool}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by dumitrel last updated on 08/Aug/21
$${x}+{y}={s};{xy}={p}\Rightarrow{p}\leqslant\frac{{s}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow \\ $$$${p}\leqslant\frac{\mathrm{1}}{\mathrm{16}}\Rightarrow{xy}\neq\frac{\mathrm{1}}{\mathrm{15}} \\ $$