Question

If x=√11+6√2 then find the value of x+1/x

Answer 1

Solution:

i) x = √11+6√2

= √11+2√9×2

=√(9+2)+2(9×2)

=√(√9)²+(√2)²+2*√9*√2

=√(√9 + √2)²

= √9 + √2

= 3+√2 ----(1)

ii) 1/x = 1/(3+√2)

= (3-√2)/[(3+√2)(3-√2)]

= (3-√2)/[3²-(√2)²

= (3-√2)/(9-2)

=(3-√2)/7 ---(2)

iii) x + 1/x

= 3+√2+(3-√2)/7

= [7(3+√2)+3-√2]/7

= (21+7√2+3-√2]/7

= (24+6√2)/7

Or

= [6(4+√2)/7]

••••

Answer 2

$x = \sqrt{11 + 6 \sqrt{2} }$

$x {}^{2} = 11 + 6 \sqrt{2}$

$x {}^{2} = {3}^{2} + ( \sqrt{2} ) {}^{2} + 2 \times 3 \times \sqrt{2} \\ x {}^{2} = (3 + \sqrt{2} ) {}^{2}$

$x = 3 + \sqrt{2}$

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

$= \frac{3 - \sqrt{2} }{(3 + \sqrt{2} )(3 - \sqrt{2)} }$

$\frac{3 - \sqrt{2} }{9 - 2} = \frac{3 - \sqrt{2} }{7}$

$x + \frac{1}{x} = 3 + \sqrt{2 } + \frac{3 - \sqrt{2} }{7}$

$= \frac{21 + 7 \sqrt{2} + 3 - \sqrt{2} }{7} {}$

$\frac{24 + 6 \sqrt{2} }{7}$