Question

if x=3+root8 find the value of x+1/x and xsquare+1/xsquare​

Answer 1

Answer:

6 and 34

Step-by-step explanation:

for this we have to use RF

x=3+2root2

1/x=1/(3+2root2). ×. (3-2root2)/(3-2root2)= (3-2root2)/(9-8)

=3-2root2

so,

x+1/x= 6

Now,

(x+1/x)^2=36

x^2+1/x^2+2x.1/x= 36

x^2+1/x^2=36-2=34

(.) represents multiplication

Answer 2

Answer:

$x + \frac{1}{x} = > 6 \\ x {}^{2} + \frac{1}{x {}^{2} } = 34$

Step-by-step explanation:

$x = 3 + \sqrt{8} = > \frac{1}{x} = \frac{1}{3 + \sqrt{8} } \\ Since, the \: denominator \\ =3+ \sqrt{8} , \\ \: it's \: rationalizing \: factor \\ =3- \sqrt{8} \: \\ Therefore, \: \frac{1}{3 + \sqrt{8} } \times \frac{3 - \sqrt{8} }{3 - \sqrt{8} } \\ = \frac{3 - \sqrt{8} }{(3) {}^{2} - ( \sqrt{8}) {}^{2} } = \frac{3 - \sqrt{8} }{9 - 8} \\ = > 3 - \sqrt{8}$

$x + \frac{1}{x} = 3 + \sqrt{8} + 3 - \sqrt{8} = > 6$

$= > (x + \frac{1}{x} ) {}^{2} = (6) {}^{2} = 36 \\ Since, \: (x + \frac{1}{x} ) {}^{2} \\ = x {}^{2} + \frac{1}{x {}^{2} } + 2 \times x \times \frac{1}{x} \\ = > x {}^{2} + \frac{1}{x {}^{2} } + 2 = 36 \\ \: \: that \: is, \: x {}^{2} + \frac{1}{x {}^{2} } = 36 - 2 \\ = > 34$