Search any question & find its solution
Question:
Answered & Verified by Expert
The product of all the real roots of $x^2-8 x+9-\frac{8}{x}+\frac{1}{x^2}=0$ is
Options:
Solution:
2854 Upvotes
Verified Answer
The correct answer is:
1
We have,
$$
\begin{aligned}
x^2-8 x+9-\frac{8}{x}+\frac{1}{x^2} & =0 \\
\Rightarrow \quad x^4-8 x^3+9 x^2-8 x+1 & =0
\end{aligned}
$$
Product of roots $=1$
$$
\begin{aligned}
x^2-8 x+9-\frac{8}{x}+\frac{1}{x^2} & =0 \\
\Rightarrow \quad x^4-8 x^3+9 x^2-8 x+1 & =0
\end{aligned}
$$
Product of roots $=1$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.