Question

If (y + 1/y) = 8 then, (y² +1/y²) is

Answer 1

$\huge\red {\bf {\underline {Solution:-}}}$

${\tt {y + \frac{1}{y} = 8}}$

$⇒{\tt { {(y + \frac{1}{y}) }^{2} = {8}^{2}}}$

$⇒{\tt { {y}^{2} + 2 \times y \times \frac{1}{y} + \frac{1}{ {y}^{2} } = 64}}$

$⇒ {\tt {{y}^{2} + 2 + \frac{1}{ {y}^{2} } = 64}}$

$⇒{\tt { {y}^{2} + \frac{1}{ {y}^{2} } = 64 - 2}}$

$⇒{\tt { {y}^{2} + \frac{1}{ {y}^{2} } = 62}}$

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Hope It Helps Uh! ❤

Answer 2

$\pink{\bf{\underline {Solution:-}}}$

${\tt{y+\frac{1}{y} = 8}}$

$\tt{⟹ {(y + \frac{1}{y}) }^{2} } = {8}^{2}$

$\tt{⟹ {y}^{2} + 2 \times y \times \frac{1}{y} + \frac{1}{ {y}^{2} } = 64}$

$\tt{⟹ {y}^{2} + 2 + \frac{1}{ {y}^{2} } = 64 }$

$\tt{⟹ {y}^{2} + \frac{1}{ {y}^{2} } = 64 - 2 }$

$\tt{⟹ {y}^{2} + \frac{1}{ {y}^{2} } = 62 }$

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$\red {\bf {\underline{Important\:Formulas:-}}}$

✷ ${(a+b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

✷ ${(a-b)}^{2} = {a}^{2} - 2ab + {b}^{2}$

✷ ${(a+b)}^{2} = {(a-b)}^{2} + 4ab$

✷ ${(a-b)}^{2} = {(a+b)}^{2} - 4ab$

✷ ${a}^{2} - {b}^{2} = (a+b)(a-b)$

✷ ${a}^{2} + {b}^{2} = {(a+b)}^{2} - 2ab$

✷ ${a}^{2} + {b}^{2} = {(a-b)}^{2} + 2ab$

✷ $4ab = {(a+b)}^{2} - {(a-b)}^{2}$

✷ $2 ({a}^{2}+{b}^{2}) = {(a+b)}^{2} + {(a-b)}^{2}$