Question

help kalo...................​

Answer 1

Given,

$\sf \quad a^{2} - \dfrac{1}{a^{2}} = 5$

To Find,

$\sf \quad a^{4} + \dfrac{1}{a^{4}} = ?$

Solution,

$\sf \rightarrow \quad {a}^{2} - \dfrac{1}{ {a}^{2} } = 5 \\ \\ \sf \quad Square \: both \: sides, \\ \\ \sf \rightarrow \quad \bigg( {a}^{2} - \frac{1}{ {a}^{2} } \bigg)^{2} = {(5)}^{2} \\ \\ \sf \rightarrow \quad {a}^{4} + \frac{1}{ {a}^{4} } - 2 \times \cancel{ {a}^{2} } \times \frac{1}{ \cancel{ {a}^{2} }} = 25 \\ \\ \sf \rightarrow \quad {a}^{4} + \frac{1}{ {a}^{4} } = 25 + 2 \\ \\ \sf \rightarrow \quad {a}^{4} + \frac{1}{ {a}^{4} } = 27$

Answer 2

$\sf{\huge{\mathfrak{\pink{\underline{Question}}}}}$

If $a^2 - \dfrac{1}{a^2} = 5$

find $a^4 -\dfrac{1}{a^4}$

$\sf{\huge{\mathfrak{\pink{\underline{solution}}}}}$

$\sf\implies a^2 - \dfrac{1}{a^2} = 5$

• squaring both the sides

$\sf\implies (a^2 - \dfrac{1}{a^2})^2 = 5^2$

$\sf{\bold{\red{\boxed{(a + b)^2 = a^2 + b^2 + 2ab }}}}$

$\sf\implies a^4 + \dfrac{1}{a^4} - 2 \times \cancel a^2 \times \dfrac{1}{\cancel a^2 } = 25$

$\sf\implies a^4 + \dfrac{1}{a^4} = 25 + 2$

$\sf\implies a^4 + \dfrac{1}{a^4} = 27$

$\sf{\underline{\boxed{\purple{\mathfrak{ a^4 + \dfrac{1}{a^4} = 27 }}}}}$

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