Question

if x + 1 / x = root 5 find the value of x² + 1/ x² and x⁴ + 1/ x⁴.​

Answer 1

EXPLANATION.

⇒ (x + 1/x) = √5.

As we know that,

Squaring on both sides of the equation, we get.

⇒ (x + 1/x)² = (√5)².

⇒ (x)² + (1/x)² + 2(x)(1/x) = 5.

⇒ (x)² + (1/x)² + 2 = 5.

⇒ x² + 1/x² = 5 - 2.

⇒ x² + 1/x² = 3. - - - - - (1).

Again squaring on both sides of the equation, we get.

⇒ [x² + 1/x²]² = (3)².

⇒ (x²)² + (1/x²)² + 2(x²)(1/x²) = 9.

⇒ x⁴ + 1/x⁴ + 2 = 9.

⇒ x⁴ + 1/x⁴ = 9 - 2.

⇒ x⁴ + 1/x⁴ = 7. - - - - - (2).

Answer 2

$\bf \huge{Given}$

$\bf{x + \frac{1}{x} = \sqrt{5} }$

$\small \text \pink{So when squaring both the sides we get }$

$\sf \red{(x + \frac{1}{x} {)}^{2} = 5}$

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

$\bf{SINCE}$

$\bf \orange{ {a}^{2} + {b}^{2} = (a + b {)}^{2} - 2ab}$

$\bf \red{ = {5 - 2} \: \: \: \: \: \: \sf{since \: (x + \frac{1}{x } {)}^{2} = 5}}$

$> > \bf \huge \boxed{3}$

$\small \text \orange{Now when we again square on both side we get}$

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

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

$\bf{SINCE}$

$\bf \orange{ {a}^{2} + {b}^{2} = (a + b {)}^{2} - 2ab}$

$\bf \green{( {3)}^{2} - 2\: \: \: \: since \: ({x}^{2} + \frac{1}{ {x}^{2} } = 3 ) }$

$\bf \purple{ = 9 - 2 }$

$> > \bf \huge \boxed{7}$

• refer the attachment