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Question Number 31306 by pieroo last updated on 05/Mar/18

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

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

Answered by MJS last updated on 05/Mar/18

(a+b)^2 =a^2 +2ab+b^2   a^2 =y^2 ⇒a=y  2ab=8y  2by=8y  b=4  (y+4)^2 =y^2 +8y+16  16=9k  k=((16)/9)

$$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} ={y}^{\mathrm{2}} \Rightarrow{a}={y} \\ $$$$\mathrm{2}{ab}=\mathrm{8}{y} \\ $$$$\mathrm{2}{by}=\mathrm{8}{y} \\ $$$${b}=\mathrm{4} \\ $$$$\left({y}+\mathrm{4}\right)^{\mathrm{2}} ={y}^{\mathrm{2}} +\mathrm{8}{y}+\mathrm{16} \\ $$$$\mathrm{16}=\mathrm{9}{k} \\ $$$${k}=\frac{\mathrm{16}}{\mathrm{9}} \\ $$

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