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Question Number 130334 by pete last updated on 24/Jan/21

Complete the square in the expression  y^2  +8y+9k and hence find the value of  k that makes it a perfect square.

$$\mathrm{Complete}\:\mathrm{the}\:\mathrm{square}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mathrm{y}^{\mathrm{2}} \:+\mathrm{8y}+\mathrm{9k}\:\mathrm{and}\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}\:\mathrm{that}\:\mathrm{makes}\:\mathrm{it}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$

Answered by TheSupreme last updated on 24/Jan/21

(y+a)=y^2 +2ay+a^2   2a=−8 →a=−4  a^2 =16=9k  k=((16)/9)

$$\left({y}+{a}\right)={y}^{\mathrm{2}} +\mathrm{2}{ay}+{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{a}=−\mathrm{8}\:\rightarrow{a}=−\mathrm{4} \\ $$$${a}^{\mathrm{2}} =\mathrm{16}=\mathrm{9}{k} \\ $$$${k}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$

Commented by pete last updated on 24/Jan/21

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Answered by john_santu last updated on 24/Jan/21

y^2 +8y+9k = (y+4)^2 +9k−16  ⇔ 9k−16 = 0 ; k = ((16)/9)

$${y}^{\mathrm{2}} +\mathrm{8}{y}+\mathrm{9}{k}\:=\:\left({y}+\mathrm{4}\right)^{\mathrm{2}} +\mathrm{9}{k}−\mathrm{16} \\ $$$$\Leftrightarrow\:\mathrm{9}{k}−\mathrm{16}\:=\:\mathrm{0}\:;\:{k}\:=\:\frac{\mathrm{16}}{\mathrm{9}} \\ $$

Commented by pete last updated on 24/Jan/21

merci

$$\mathrm{merci} \\ $$

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