if a+1/a=3 then a^2+1/a^2=?
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[tex] \sf a + \frac{1}{a} = 3[/tex]
Squaring both sides, we get,
[tex] \sf {(a + \frac{1}{a}) }^{2} = {3}^{2} [/tex]
[tex] \sf \implies {a}^{2} + \frac{1}{ {a}^{2} } + 2 \times \cancel{a} \times \frac{1}{ \cancel{a}} = 9[/tex]
[tex] \sf \implies {a}^{2} + \frac{1}{ {a}^{2} } + 2 = 9[/tex]
[tex] \sf \implies {a}^{2} + \frac{1}{ {a}^{2} } = 9 - 2[/tex]
[tex] \sf \implies {a}^{2} + \frac{1}{ {a}^{2} } = 7[/tex]
Hence, the value of a²+1/a² is 7.
[tex]\huge\mathbb\pink{\underline{\underline{Answer}}}[/tex]
[tex]a + \frac{1}{a} = 3[/tex]
To find [tex]\longrightarrow[/tex]
[tex] {a}^{2} + \frac{1}{ {a}^{2} } [/tex]
[tex]a + \frac{1}{a} = 3[/tex]
Squaring both the sides.
[tex] (a + \frac{1}{a} )^{2} = {3}^{2} [/tex]
[tex] {a }^{2} + 2 \times a \times \frac{1}{a} + { \frac{1}{a} }^{2} = 9 [/tex]
[tex] {a}^{2} + 2 + \frac{1}{ {a}^{2} } = 9 [/tex]
[tex] {a}^{2} + \frac{1}{ {a}^{2} } = 9 - 2[/tex]
[tex] {a}^{2} + \frac{1}{ {a}^{2} } = 7[/tex]