Question Number 145057 by 7770 last updated on 02/Jul/21
$$\boldsymbol{{find}}\:\:\boldsymbol{{k}}\:\:\boldsymbol{{if}}\:\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}+\boldsymbol{{k}}\:\boldsymbol{{is}}\:\boldsymbol{{a}}\:\boldsymbol{{perfect}} \\ $$$$\boldsymbol{{square}} \\ $$
Commented by mr W last updated on 02/Jul/21
$${y}^{\mathrm{2}} +{y}+\frac{\mathrm{1}}{\mathrm{4}}=\left({y}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by 7770 last updated on 02/Jul/21
$$\boldsymbol{{s}}{orry}\:{its}\:{y}^{\mathrm{2}} +{y}+{k}\:{is}\:{a}\:{perfect}\:{square} \\ $$
Commented by imjagoll last updated on 02/Jul/21
$$\mathrm{y}^{\mathrm{2}} +\mathrm{y}+\mathrm{k}\:\rightarrow\mathrm{perfect}\:\mathrm{square}\:\mathrm{if}\:\bigtriangleup=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{4k}=\mathrm{0}\:;\:\mathrm{k}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$