If x = sqrt{7 + 4sqrt{3}},then x + frac{1}{x} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x = \sqrt{7 + 4\sqrt{3}}$.We can simplify $x$ by writing $7 + 4\sqrt{3}$ as $(2 + \sqrt{3})^2$:$x = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given $x = \sqrt{7 + 4\sqrt{3}}$.We can simplify $x$ by writing $7 + 4\sqrt{3}$ as $(2 + \sqrt{3})^2$:$x = \sqrt{(2 + \sqrt{3})^2} = 2 + \sqrt{3}$.Now,find $\frac{1}{x}$:$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} 	imes \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$.Therefore,$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.

If $x = \sqrt{7 + 4\sqrt{3}}$,then $x + \frac{1}{x} = ......$

Similar Questions

The least number among $\sqrt[3]{4}, \sqrt[4]{5}, \sqrt[4]{7}$ and $\sqrt[3]{8}$ is:

If $3^x - 3^{x - 1} = 6$,then $x^x$ is equal to

If $\frac{{({2^{n + 1}})^m}({2^{2n}}){2^n}}{{({2^{m + 1}})^n}{2^{2m}}} = 1$,then $m =$

The number of digits in the decimal expansion of $16^5 \times 5^{16}$ is

The rationalizing factor of $2\sqrt{3} - \sqrt{7}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module