If \((2+i)\) is a root of the equation \(x^3-5 x^2+9 x-5=0\), then the other roots are
If \((2+i)\) is a root of the equation \(x^3-5 x^2+9 x-5=0\), then the other roots are
1 and \((2-i)\)
-1 and \((3+i)\)
0 and 1
-1 and \((-2+i)\)
Solution
It is given that \(2+i\) is the root of the equation \(x^3-5 x^2+9 x-5=0\), so another non-real complex root will be \(2-i\).
Now, let the third root is \(\alpha\), so by product of roots, we have
\((2+i)(2-i) \alpha=5 \Rightarrow \alpha=1\)
Hence, option (a) is correct.