Question

plzz guys❤️❤️❤️...... ​

Answer 1

SOLUTION

Given,

x= 2+ √3

So, x^2 = (2+√3)^2

=) (2)^2 + (√3)^2 + 2× 2×√3

=) 4+ 3+ 4√3

=) 7+ 4√3

Now,

$= > \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \\ = > \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } = \frac{2 - \sqrt{3} }{4 - 3} \\ = > 2 - \sqrt{3} \\ \\ so \\ = > ( \frac{1}{x} ) {}^{2} = (2 - \sqrt{3} ) {}^{2} = 4 + 3 - 4 \sqrt{3} \\ = > 7 - 4 \sqrt{ 3} \\ \\ = > {x}^{2} + \frac{1}{x {}^{2} } = 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} \\ = > 14$

hope it helps ✔️

Answer 2

Given,

x=2+√3

So,x^2=(2+√3)^2

=)(2)^2+(√3)^2+2×2×√3

=)4+3+4√3

=)7+4√3

Now,

=  >  \frac{1}{x}  =  \frac{1}{2 +  \sqrt{3} }  \\  =  >  \frac{1}{2 +  \sqrt{3} }  \times  \frac{2 -  \sqrt{3} }{2 -  \sqrt{3} }  =  \frac{2 -  \sqrt{3} }{4 - 3}  \\  =  > 2 -  \sqrt{3}  \\  \\ so \\   =  >  ( \frac{1}{x} ) {}^{2}  = (2 -  \sqrt{3} ) {}^{2}  = 4 + 3 - 4 \sqrt{3}  \\  =  > 7 - 4 \sqrt{ 3}  \\  \\  =  >  {x}^{2}  +  \frac{1}{x {}^{2} }  = 7 + 4 \sqrt{3}  + 7 - 4  \sqrt{3}  \\  =  > 14