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

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

Answered by TheSupreme last updated on 24/Jan/21

$$(y+a)=y^2+2ay+a^2$$$$2a=-8\to a=-4$$$$a^2=16=9k$$$$k=\frac{16}{9}$$

Commented by pete last updated on 24/Jan/21

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

Answered by john_santu last updated on 24/Jan/21

$$y^2+8y+9k={(y+4)}^2+9k-16$$$$\iff 9k-16=0;k=\frac{16}{9}$$

Commented by pete last updated on 24/Jan/21

$$\mathrm{merci}$$

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