MCQ
If $\text{x}+\frac{1}{\text{x}}=4,$ then $\text{x}^4+\frac{1}{\text{x}^4}=$
- A$169$
- B$190$
- ✓$194$
- D$192$
✓
Answer
Correct option: C.
$194$
Using, $(a+b)^2=a^2+b^2+2 a b$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\frac{1}{\text{x}}$
$\Rightarrow(4)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=16-2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=16-2$
$\text{x}^2+\frac{1}{\text{x}^2}=16-2$
Again, squaring both sides,
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow\text{x}^4+\Big(\frac{1}{\text{x}^4}+2\text{x}^2\times\frac{1}{\text{x}^2}=196\Big)$
$\Rightarrow\text{x}^4+\Big(\frac{1}{\text{x}^4}\Big)=196-2$
$\Rightarrow\text{x}^4+\Big(\frac{1}{\text{x}^4}\Big)=194$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2\times\text{x}\frac{1}{\text{x}}$
$\Rightarrow(4)^2=\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)+2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=16-2$
$\Rightarrow\text{x}^2+\Big(\frac{1}{\text{x}^2}\Big)=16-2$
$\text{x}^2+\frac{1}{\text{x}^2}=16-2$
Again, squaring both sides,
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$
$\Rightarrow\text{x}^4+\Big(\frac{1}{\text{x}^4}+2\text{x}^2\times\frac{1}{\text{x}^2}=196\Big)$
$\Rightarrow\text{x}^4+\Big(\frac{1}{\text{x}^4}\Big)=196-2$
$\Rightarrow\text{x}^4+\Big(\frac{1}{\text{x}^4}\Big)=194$
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 median of a triangle divides it into two:
If $x^{-2}=64$, then $x^{1 / 3}+x^0=$
→The ordinate of any point on x-axis is
In fig. ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, $\angle B A D=$
→The congruence rule, by which the two triangles in the given figure are congruent is$............$
→Chords $AD$ and $BC$ intersect each other at right angles at point $P. \angle\text{DAB}=35^\circ,$ then $\angle\text{ADC}$ is equal to:
→In the given figure, $O$ is the centre of the circle $ABE$ is a straight line. if $\angle\text{DBE}=95^\circ$ then $\angle\text{AOD}$ is equal to:
→In Fig. O is the centre of the circle and $\angle B D C=42^{\circ}$. The measure of $\angle A C B$ is
→If $\bar{\text{x}_1},\bar{\text{x}}_2,\ _{\dots},\bar{\text{x}}_\text{n}$ are the means of $n$ group with $\mathrm{n}_1, \mathrm{n}_2, \ldots, \mathrm{n}_{\mathrm{n}}$ number of observation respectively then the mean $\bar{\text{x}}$ of all the groups taken together is:
→If $p(x) = x + 3$, then $p(x) + p(-x)$ is equal to:
→