Question

if x=2+√3. Find x-1/xand x²+1/x²

Answer 1

Hey mate!!

$x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} = 2 - \sqrt{3}$

Now,

$x - \frac{1}{x} = \cancel 2 + \sqrt{3} - \cancel 2 + \sqrt{3} \\ \\ \bold{\boxed{\red{ x - \frac{1}{x} = 2 \sqrt{3}} }}$

Again,

$x - \frac{1}{x} = 2 \sqrt{3} \\ \\ (x - \frac{1}{x} ) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } - 2 \times \cancel x \times \frac{1}{ \cancel x} \\ \\ (2 \sqrt{3} ) {}^{2} = x {}^{2} + \frac{1}{x {}^{2} } - 2 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 12 +2 \\ \\ \bold{\boxed{\red{ x {}^{2} + \frac{1}{x {}^{2} } = 14 }}}$

Thanks for the question!

Answer 2

x=2+root3
1/x =1/2+root3
rationalise denominator
1/2+root3 × 2-root3/2-root3
1/x = 2-root3/(2)^2- (root3)^2
1/x =2-root3/4-3
1/x = 2-root3/1
1/x=2-root3
given
x-1/x
(2+root3) - (2-root3)
2+root3-2+root3
root3+root3
=2root3
x-1/x =2root3 is answer obtained

now to find x^2 + 1/x^2 take
x-1/x =2root3
squaring on both sides
(x-1/x)^2 =(2root3) ^2
it is of form (a-b) ^2 =a^2+b^2-2ab
x^2 +1/x^2 -2(x)(1/x) =4×3
x^2 +1/x^2 -2=12
x^2 +1/x^2=12+2
x^2 +1/x^2=14 is answer obtained