If $$ x=2+\sqrt{3} $$, find the value of $$ x^{2}+\frac{1}{x^{2}} $$

Question

UpStudy AI · Accepted Answer

Let $$ x = 2+\sqrt{3} $$. First, find $$ \frac{1}{x} $$:

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

Multiply the numerator and denominator by the conjugate $$ 2-\sqrt{3} $$:

$$
\frac{1}{2+\sqrt{3}} = \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}.
$$

The denominator is a difference of squares:

$$
(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1.
$$

Therefore,

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

Now, compute $$ x + \frac{1}{x} $$:

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

Recall the identity:

$$
\left(x+\frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2},
$$

which can be rearranged to find $$ x^2+\frac{1}{x^2} $$:

$$
x^2+\frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 - 2.
$$

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

$$
x^2+\frac{1}{x^2} = 4^2 - 2 = 16 - 2 = 14.
$$

Thus, the value of $$ x^2+\frac{1}{x^2} $$ is $$ 14 $$.
$$ x^{2}+\frac{1}{x^{2}} = 14 $$

If \( x=2+\sqrt{3} \), find the value of \( x^{2}+\frac{1}{x^{2}} \)