Question

If (a^2+1/a^2)=14 then
(a+1/a)=?

Answer 1

Answer:

GIVEN, a² + 1/a² = 14.

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We know, a²+b² = (a+b)² - 2 ab

So, (a+1/a) ² - 2 = 14

=> (a+1/a) = √16 = 4

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We also know, a²+b² = (a-b)² + 2 ab

So, (a-1/a)² + 2 = 14

=> (a - 1/a) = √12

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Ultimately, a² - 1/a²

= (a + 1/a) (a - 1/a)

= 4√12

This is your answer.

Answer 2

GIVEN :-

$\\ \sf \: \left( {a}^{2} + \dfrac{1}{ {a}^{2} } \right) = 14$

TO FIND :-

$\\ \sf \: a + \dfrac{1}{a}$

FORMULA USED :-

$\\ \bigstar\boxed{\sf \: (x + {y)}^{2} = {x}^{2} + {y}^{2} + 2xy} \\ \\ \therefore \boxed{ \sf \: {x}^{2} + {y}^{2} = (x + {y)}^{2} - 2xy}$

SOLUTION :-

We know , x² + y² = (x+y)² - 2xy

In the above question ,

• x = a
• y = 1/a

Substituting values in the formula ,

$\\ \implies\sf \: {a}^{2} + \dfrac{1}{ {a}^{2} } = { \left( a + \dfrac{1}{a} \right)}^{2} - 2( \cancel{a}) \left( \dfrac{1}{ \cancel{a}} \right) \\ \\ \\ \implies\sf \: 14 = { \left(a + \dfrac{1}{a} \right)}^{2} - 2 \\ \\ \\ \implies \sf \: { \left(a + \dfrac{1}{a} \right)}^{2} = 14 + 2$

$\\ \implies\sf \: { \left(a + \dfrac{1}{a} \right)}^{2} = 16$

Taking root , we get...

$\\ \implies\sf \: a + \dfrac{1}{a} = \sqrt{16} \\ \\ \\ \implies\boxed{ \bf \: a + \dfrac{1}{a} = 4 }$

MORE IDENTITIES :-

$\boxed{ \tt\:(x - {y})^{2} = {x}^{2} + {y}^{2} - 2xy} \\ \\ \boxed{ \tt \: (x + y)(x - y) = {x}^{2} - {y}^{2} } \\ \\ \boxed{ \tt \: (x + a)(x + b) = {x}^{2} + (a + b)x + ab }$