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

The Correct Option is A

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) : \(34\sqrt5\).