Question

if x is equal to 2 + root 3 find the value of x square + 1 by x square​

Answer 1

If x=2+√3

then

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

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

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

$\frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}$

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

Hence,

$= > x + \frac{1}{x}$

$= > 2 + \sqrt{3} + 2 - \sqrt{3}$

$= > 4$

Now,if

$x + \frac{1}{x} = 4$

Squaring both the sides-

$(x + \frac{1}{x} ) {}^{2} = (4) {}^{2}$

${x}^{2} + \frac{1}{ {x}^{2} } + 2. \frac{1}{x} .x = 16$

Hence,

${x}^{2} + \frac{1}{ {x}^{2} } = 14$