Question

If a^2-5a+1=0, find the value of a^2 + 1/a^2

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Answer 1

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$\large{\underbrace{\sf{\purple{required \: solution:}}}}$

$\longrightarrow \sf \purple{23}$

Given that,

• a² - 5a + 1 = 0

$\longrightarrow \boxed{ \purple {\sf Equation \leadsto {a}^{2} + \dfrac{1}{ {a}^{2} } }}$

$\rightarrow \sf {a}^{2} - 5a + 1$

$\rightarrow \sf {a}^{2} + 1 = 5a$

$\rightarrow \sf \dfrac{a + 1}{a} = 5$

$\rightarrow \sf \dfrac{ {a}^{2} }{a} + \dfrac{1}{a} = 5$

$\rightarrow \sf \dfrac{ \cancel{{a}^{2} }}{ \cancel{a}} + \dfrac{1}{a} = 5$

$\rightarrow \sf a+ \dfrac{1}{a} = 5$

On squaring both sides , we have :

$\rightarrow \sf {(a + \dfrac{a}{1} )}^{2} = {5}^{2}$

$\rightarrow \sf {a}^{2} + \dfrac{1}{ {a}^{2} } = \: ?$

$\rightarrow \sf {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times a \times \dfrac{1}{a} = 25$

$\rightarrow \sf {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 \times \cancel{ a} \times \dfrac{1}{ \cancel{a}} = 25$

$\rightarrow \sf {a}^{2} + \dfrac{1}{ {a}^{2} } + 2 = 25$

$\rightarrow \sf {a}^{2} + \dfrac{1}{ {a}^{2} } = 25 - 2$

$\rightarrow \boxed{ \purple{\sf {a}^{2} + \dfrac{1}{ {a}^{2} } = 23}}$

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Answer 2

To find a^2 + 1/a^2

a^2-5a+1=0

a^2 +1 =5a

Diving equation by "a"

$\frac{a ^{2} }{a} + \frac{1}{ a} = 5$

$a + \frac{1}{a} = 5$

Squaring both sides,

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

$(a) ^{2} + ( \frac{1}{a^{} } )^{2} + 2 \times a \times \frac{1}{a} = 25$

${a}^{2} + \frac{1}{a ^{2} } + 2 = 25$

${a}^{2} + \frac{1}{a ^{2} } = 25 - 2 = 23$

Hence, 23 is the answer