Question

evaluate the question given in attachment​

Answer 1

Given :-

$x + \frac{1}{x} = 4$

To find :-

$x {}^{2} + \frac{1}{x {}^{2} } =$

Formula Implemented

( a + b )² = a² + b² + 2ab

Lets do!

For given equation We have to do Squaring on both sides

Applying (a + b )² = a² + b² + 2ab formula

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

$(4) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } + 2$

Tranpose 2 to RHS

$x {}^{2} + \frac{1}{x {}^{2} } = 16-2$

$x {}^{2} + \frac{1}{x {}^{2} } = 14$

This is the required answer

Know more Some algebraic identities :-

a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

Answer:

Given :-

x + \frac{1}{x} = 4x+

x

1

=4

To find :-

x {}^{2} + \frac{1}{x {}^{2} } =x

2

+

x

2

1

=

Formula Implemented

( a + b )² = a² + b² + 2ab

Lets do!

For given equation We have to do Squaring on both sides

Applying (a + b )² = a² + b² + 2ab formula

(x + \frac{1}{x} ) {}^{2} = x {}^{2} + \frac{1 {}^{2} }{x {}^{2} } + 2 \times x \times \frac{1}{x}(x+

x

1

)

2

=x

2

+

x

2

1

2

+2×x×

x

1

(4) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } + 2(4)

2

=x

2

+

x

2

1

+2

Tranpose 2 to RHS

x {}^{2} + \frac{1}{x {}^{2} } = 16-2x

2

+

x

2

1

=16−2

x {}^{2} + \frac{1}{x {}^{2} } = 14x

2

+

x

2

1

=14

This is the required answer

Know more Some algebraic identities :-

a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc