Question

please answer.

x=√2+6.Find x+1/x.

Answer 1

Answer:

(33√2+210)/34

Step-by-step explanation:

x=√2+6

1/x=1/√2+6

=1/√2+6×√2-6/√2-6

=√2-6/(√2)^2-(6)^2

=√2-6/2-36

=√2-6/-34

=6-√2/34

x+1/x=√2+6+6-√2/34

=(34√2+204-√2+6)/34

=(33√2+210)/34

Answer 2

Answer :-

$\tt x + \dfrac{1}{x} = \dfrac{ 33\sqrt{2} + 210 }{34}$

Solution :-

x = √2 + 6

1/x = 1/(√2 + 6)

Rationalise the denominator

$\sf \dfrac{1}{x} = \dfrac{1}{ \sqrt{2} + 6}$

The rationalising factor √2 + 6 is √2 - 6. So multiply both numerator and denominator.

$\sf = \dfrac{1}{ \sqrt{2} + 6} \times \dfrac{ \sqrt{2} - 6 }{ \sqrt{2} - 6 }$

$\sf = \dfrac{1( \sqrt{2} - 6) }{ \sqrt{2} + 6( \sqrt{2} - 6)}$

$\sf = \dfrac{ \sqrt{2} - 6}{ \sqrt{2} + 6( \sqrt{2} - 6)}$

$\sf = \dfrac{ \sqrt{2} - 6}{ {( \sqrt{2})}^{2} - {(6) }^{2} }$

[Because (x + y)(x - y) = x² - y²]

$\sf = \dfrac{ \sqrt{2} - 6}{2 - 36 }$

$\sf = \dfrac{ \sqrt{2} - 6}{ - 34}$

$\sf = - \dfrac{ \sqrt{2} - 6}{34}$

$\sf \dfrac{1}{x} = - \dfrac{ \sqrt{2} - 6 }{34}$

Now find the value of x + 1/x

$\sf x + \dfrac{1}{x} = \sqrt{2} + 6 + ( - \dfrac{ \sqrt{2} - 6 }{34})$

$\sf = \sqrt{2} + 6 - \dfrac{ \sqrt{2} - 6 }{34}$

Taking LCM

$\sf = \dfrac{ \sqrt{2}(34)}{1(34)} + \dfrac{6(34)}{1(34)} - \dfrac{ \sqrt{2} - 6 }{34}$

$\sf = \dfrac{ 34\sqrt{2}}{34} + \dfrac{204}{34} - \dfrac{ \sqrt{2} - 6 }{34}$

$\sf = \dfrac{ 34\sqrt{2} + 204 - (\sqrt{2} - 6)}{34}$

$\sf = \dfrac{ 34\sqrt{2} + 204 - \sqrt{2} + 6}{34}$

$\sf = \dfrac{ 33\sqrt{2} + 210 }{34}$

$\bf x + \dfrac{1}{x} = \dfrac{ 33\sqrt{2} + 210 }{34}$