Question

If $x$ is a positive real number such that $x^8+\bigg(\frac{1}{x}\bigg)^8=47$ , then the value of $x^9+\bigg(\frac{1}{x}\bigg)^9$ is

• A. $34\sqrt 5$
• B. $40\sqrt5$
• C. $36\sqrt5$
• D. $30\sqrt5$

Answer 1

Answer: (A)

$34\sqrt 5$

Answer 2

Given :
$x^8+(\frac{1}{8})^8=47$, this can be also expressed as :

⇒ $(x^4)^2+(\frac{1}{x^4})^2=47$

⇒ $(x^4+\frac{1}{x^4})^2-2\times x^4\times\frac{1}{x^4}=47$

⇒ $(x^4+\frac{1}{x^4})^2=49$

⇒ $x^4+\frac{1}{x^4}=7$ …… (i)

Now , the above equation (i) can be expressed as :
⇒ $(x^2)^2+(\frac{1}{x^2})^2=7$

⇒ $(x^2+\frac{1}{x^2})-2\times x^2\times \frac{1}{x^2}=7$

⇒ $(x^2+\frac{1}{x^2})^2=9$ i.e $x^2+\frac{1}{x^2}=3$

Similarly , we get :
$x+\frac{1}{x}=\sqrt5$
⇒ $x^3+\frac{1}{x^3}=(x+\frac{1}{x})^3-3\times x\times \frac{1}{x}(x+\frac{1}{x})$
$x^3+\frac{1}{x^3}=(\sqrt5)^3-3\sqrt5=2\sqrt5$

Similarly like the above, we can say that :
⇒ $x^9+\frac{1}{x^9}=(x^3+\frac{1}{x^3})^3-3\times x \times \frac{1}{x^3}(x^3+\frac{1}{x^3})$

⇒ $x^9+\frac{1}{x^9}=(2\sqrt5)^3-3(2\sqrt5)$

⇒ $x^9+\frac{1}{x^9}=40\sqrt5-6\sqrt5$
$=34\sqrt5$
Therefore, the correct option is (A) : $34\sqrt5$.