Question

If
${a}^{2} - 5a \: + \: 1$
find,

i)
$a \: + \: \frac{1}{a}$
ii)
${a}^{2} \: + \: \frac{1}{ {a}^{2} }$
​

Answer 1

$\big a+\frac{1}{a}=\sqrt{29}$

$\big a^{2}-\frac{1}{a^{2}=5\sqrt{29}}$

$\bigstar\underline\bold{Given}$

$\color{orchid}\underline{\underline{ \tt{{ \boxed {Equation\: a^{2}-5a-1=0a} }}}}$

$\bigstar\underline\bold{To\: find}$

$\color{orange} a+\frac{1}{a},a^{2}-\frac{1}{a^2}$

$\bigstar\underline\bold{Solution}$

$\color{violet}a(a-\frac{1}{a})=5$

$\color{violet}a-\frac{1}{a}=5$

$\bigstar\underline\bold{Taking,}$

$\color{darkgreen}(a+\frac{1}{a})^2=(a-\frac{1}{a})^2+4a\times \frac{1}{a}$

$\color{teal}(a+\frac{1}{a})^2=(5)^2+4$

$\color{teal}(a+\frac{1}{a})^2=25+4$

$\color{teal}(a+\frac{1}{a})^2=29$

$\small\textbf{Taking root both side,}$

$\color{blue}a+\frac{1}{a}=\sqrt{29}$

$\bigstar\underline\bold{Taking,}$

$\color{red}a^{2}-\frac{1}{a^{2}=(a+\frac{1}{a})(a-\frac{1}{a}})$

$\color{red}a^2-\frac{1}{a^2}=(\sqrt{29})(5)$

$\color{red}a^2-\frac{1}{a^2}=5\sqrt{29}$

Thankyou :)

Answer 2

Answer:

Annyeonghaseyo army

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