Question

if
$x + \frac{1}{x} = \sqrt{5}$
then find the value of
${x}^{2} + \frac{1}{ {x}^{2} }$
and
${x}^{4} + \frac{1}{ {x}^{4} }$
​

Answer 1

Answer:

$x^2+\frac{1} {x^{2} }=3$

$x^4+\frac{1} {x^{4} }=7$

Step-by-step explanation:

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

⇒$(\sqrt{5})^{2} =x^2+\frac{1} {x^{2} } +2$

⇒${5} =x^2+\frac{1} {x^{2} } +2$

⇒$x^2+\frac{1} {x^{2} }=3$

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

⇒$(3 )^{2} =x^4+\frac{1} {x^{4} } +2$

⇒$9=x^4+\frac{1} {x^{4} } +2$

⇒$x^4+\frac{1} {x^{4} }=7$