MCQ
If $\text{x}=3+\sqrt{8},$ than the value of $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$ is:
- A32
- B34
- C6
- D12
✓
Answer
- 34
Solution:
Given $\text{x}=3+\sqrt{8},$
$\frac{1}{\text{x}}=\frac{1}{(3+\sqrt{8})}=\frac{1}{(3+\sqrt{8})}\times\frac{(3-\sqrt{8})}{(3-\sqrt{8})}$
$=\frac{(3-\sqrt{8})}{(3^2-(\sqrt{8})^2}=\frac{(3+\sqrt{8})}{(9-8)}=(3-\sqrt{8})$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=(3+\sqrt{8})+(3-\sqrt{8})=6$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=6^2=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big) =36-2=34$
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 $x-a$ is a factor of $x^3-3 x^2 a+2 a^2 x+b$, then the value of $b$ is
→→
In $\triangle\text{AOC}$ and $\triangle\text{XYZ},\ \angle\text{A}=\angle\text{X},\ \angle\text{AO} = \angle\text{XY},\ \angle\text{AC} = \angle\text{XZ},$ then by which congruence rule $\triangle\text{AOC}\cong\triangle\text{XYZ}?$
→In $\triangle\text{PQR},\ \angle\text{R} = \angle\text{P} $ and QR = 4cm and PR = 5cm. Then the length of PQ is:
→In a $\triangle\text{ABC},$ if $\angle\text{A}=60^\circ,\angle\text{B}=80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at O, then $\angle\text{BOC}=$
→A linear equation in two variables is of the form ax + by + c = 0, where?
If $\frac{x}{y}+\frac{y}{x}=-1(x, y \neq 0)$, the value of $x^3-y^3$ is
→The area of quadrilateral PQRS, in which PQ = 7cm, QR = 6cm, RS = 12cm, PS = 15cm and PR = 9cm:
If O(0, 0), A(3, 0), B(3, 4), C(0, 4) are four given points then the figure OABC is a:
- Square.
- Rectangle.
- Trapezium.
- Rhombus.
If the degree measures of the angles of quadrilateral are 4x, 7x, 9x and 10, what is the sum of the measures of the smallest angle and largest angle?
- 140°
- 150°
- 168°
- 180°