Question
If $\text{x}=2+\sqrt{3}$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ equals:
- $-2{\sqrt{3}}$
- 2
- 4
- $4-2\sqrt{3}$
✓
Answer
- 4
Solution:
$\text{x}=2+\sqrt{3}$
$\therefore\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4$
Hence, the correct option is (c).
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The sides of a triangle are 50cm, 78cm and 112cm. The smallest altitude is:
In the given figure, ABCD is a || gm and E is the mid-point of BC. Also, DE and AB when produced meet at F. Then:
→- $\text{AF}=\frac{3}{2}\text{AB}$
- AF = 2AB
- AF = 3AB
- AF2 = 2AB2
The value of $64^{-\frac{1}{3}}\Big(64^{\frac{1}{3}}-64^{\frac{2}{3}}\Big)$ is:
→If the diagonals of a quadrilateral bisect each other, and opposite sides are parallel and equal, then the quadrilateral must be.
The length of each side of an equilateral triangle having an area of $9\sqrt{3}\text{cm}^2$ is:
→It is known that if x + y = 10 then x + y + z = 10 + z. The Euclid’s axiom that illustrates this statement is:
The base BC of triangle ABC is produced both ways and the measure of exterior angles formed are 94º and 126º. Then, $\angle\text{BAC} =$
→In $\triangle\text{ABC, }\angle\text{C}=\angle\text{A},$ BC = 4cm and AC = 5cm. then, AB =?
→- 4cm
- 5cm
- 8cm
- 2.5cm
If 'm' is a positive integer which is not a perfect square, then $\sqrt{\text{m}}$ is:
→The simplest form of $0.12\overline{3}$ is:
→- $\frac{41}{330}$
- $\frac{37}{330}$
- $\frac{41}{333}$
- None of these.