Question Number 130334 by pete last updated on 24/Jan/21
$$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{y}^{\mathrm{2}} \:+\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$
Answered by TheSupreme last updated on 24/Jan/21
$$\left({y}+{a}\right)={y}^{\mathrm{2}} +\mathrm{2}{ay}+{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}=−\mathrm{8}\:\rightarrow{a}=−\mathrm{4} \\ $$$${a}^{\mathrm{2}} =\mathrm{16}=\mathrm{9}{k} \\ $$$${k}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$
Commented by pete last updated on 24/Jan/21
$$\mathrm{thanks}\:\mathrm{sir} \\ $$
Answered by john_santu last updated on 24/Jan/21
$${y}^{\mathrm{2}} +\mathrm{8}{y}+\mathrm{9}{k}\:=\:\left({y}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{9}{k}−\mathrm{16} \\ $$$$\Leftrightarrow\:\mathrm{9}{k}−\mathrm{16}\:=\:\mathrm{0}\:;\:{k}\:=\:\frac{\mathrm{16}}{\mathrm{9}} \\ $$
Commented by pete last updated on 24/Jan/21
$$\mathrm{merci} \\ $$