MCQ
If $x=7-4 \sqrt{3}$, then $x+\frac{1}{x}=$
- A$8 \sqrt{3}$
- B49
- C48
- ✓14
✓
Answer
Correct option: D.
14
(d)
We have, $x=7-4 \sqrt{3}$.
$\therefore \quad \frac{1}{x}=\frac{1}{7-4 \sqrt{3}}=\frac{7+4 \sqrt{3}}{(7+4 \sqrt{3})(7-4 \sqrt{3})}=\frac{7+4 \sqrt{3}}{49-48}=7+4 \sqrt{3}$
$x+\frac{1}{x}=7-4 \sqrt{3}+7+4 \sqrt{3}=14$
We have, $x=7-4 \sqrt{3}$.
$\therefore \quad \frac{1}{x}=\frac{1}{7-4 \sqrt{3}}=\frac{7+4 \sqrt{3}}{(7+4 \sqrt{3})(7-4 \sqrt{3})}=\frac{7+4 \sqrt{3}}{49-48}=7+4 \sqrt{3}$
$x+\frac{1}{x}=7-4 \sqrt{3}+7+4 \sqrt{3}=14$
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
If the point $(3, 4)$ lies on the graph of $3y = ax + 7$ then the value of a is:
→In the given figure, $ABCD$ is a Rectangle. Find the values of $x$ and $y?$
→Area of a right-angled triangle is $6 cm^2$ and its perimeter is 12 cm. Then length of its hypotenuse, is
→Between any two rational numbers there:
$9^3+(-3)^3-6^3=?$
→The equation of $y-$axis is:
→The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a:
In Fig. which of the following statements must be true?
(i) a+b=d+c$\quad$(ii) $a+c+e=180^{\circ}$$\quad$(iii) b+f=c+e
→(i) a+b=d+c$\quad$(ii) $a+c+e=180^{\circ}$$\quad$(iii) b+f=c+e
The height of a cone is $21\ cm$ and its slant height is $28\ cm.$ The volume of the cone is.
→The co-ordinates of a point below the $x-$axis lying on $y-$axis at a distance of $4$ units are.
→