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Question Number 212327 by MASANJAJJ last updated on 10/Oct/24
if (a+(1/a))=15 find the value of (a^2  +(1/a^2 ))
$${if}\:\left({a}+\frac{\mathrm{1}}{{a}}\right)=\mathrm{15}\:{find}\:{the}\:{value}\:{of}\:\left({a}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right) \\ $$
Answered by universe last updated on 10/Oct/24
(a+(1/a))^2  = 15^2   a^2 +(1/a^2 ) +2 = 225  a^2 +(1/a^2 ) = 225−2 = 223
$$\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} \:=\:\mathrm{15}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:+\mathrm{2}\:=\:\mathrm{225} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:=\:\mathrm{225}−\mathrm{2}\:=\:\mathrm{223} \\ $$
Answered by mehdee7396 last updated on 10/Oct/24
(a+(1/a))^2 =225⇒a^2 +(1/a^2 )=223 ✓
$$\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} =\mathrm{225}\Rightarrow{a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\mathrm{223}\:\checkmark \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 10/Oct/24
Another way  a+(1/a)=15 ; a^2 +(1/a^2 )=?  a+(1/a)=15⇒a^2 =15a−1  ⇒(1/a)=15−a⇒(1/a^2 )=a^2 −30a+225  =15a−1−30a+225=224−15a     a^2 +(1/a^2 )=(15a−1)+(224−15a)=223
$$\mathrm{Another}\:\mathrm{way} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{15}\:;\:{a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=? \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\mathrm{15}\Rightarrow{a}^{\mathrm{2}} =\mathrm{15}{a}−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}=\mathrm{15}−{a}\Rightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }={a}^{\mathrm{2}} −\mathrm{30}{a}+\mathrm{225} \\ $$$$=\mathrm{15}{a}−\mathrm{1}−\mathrm{30}{a}+\mathrm{225}=\mathrm{224}−\mathrm{15}{a} \\ $$$$\: \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{{a}^{\mathrm{2}} }=\left(\mathrm{15}{a}−\mathrm{1}\right)+\left(\mathrm{224}−\mathrm{15}{a}\right)=\mathrm{223} \\ $$

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