If $\alpha, \beta$ are the irrational roots of the equation $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$, then the roots…
If $\alpha, \beta$ are the irrational roots of the equation $x^5-5 x^4+9 x^3-9 x^2+5 x-1=0$, then the roots of the equation $(\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0$ are
$-1, \frac{1}{3}$
$\frac{3 \pm \sqrt{5}}{2}$
$\frac{1 \pm i \sqrt{3}}{2}$
$1,-\frac{1}{3}$
Solution
Given equation,
$
x^5-5 x^4+9 x^3-9 x^2+5 x-1=0
$
$x=1$ is one root of equation.
So, $(x-1)\left(x^4-4 x^3+5 x^2-4 x+1\right)=0$
$
\Rightarrow \quad x^4-4 x^3+5 x^2-4 x+1=0
$
On dividing by $x^2$, we get,
$
\begin{aligned}
& \Rightarrow \quad x^2-4 x+5-\frac{4}{x}+\frac{1}{x^2}=0 \\
& \Rightarrow \quad\left(x^2+\frac{1}{x^2}\right)-4\left(x+\frac{1}{x}\right)+5=0 \\
& \Rightarrow \quad\left(x+\frac{1}{x}\right)^2-4\left(x+\frac{1}{x}\right)+3=0 \\
& {\left[\text { as } x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2\right]}
\end{aligned}
$
Now, let $x+\frac{1}{x}=y$
$
\begin{aligned}
y^2-4 y+3 & =0 \\
(y-1)(y-3) & =0 \\
y & =1,3 \\
\Rightarrow \quad x+\frac{1}{x}=1 \text { and } x+\frac{1}{x} & =3 . \\
\Rightarrow \quad & \quad x^2+1=x \text { and } x^2+1=3 x \\
\Rightarrow \quad &
\end{aligned}
$
Equation $x^2-3 x+1=0$ gives irrational roots
Let $\alpha, \beta$ are roots then,
$\begin{aligned} & \alpha+\beta=3, \alpha \beta=1 \text { putting these values in } \\ & (\alpha+\beta) x^2+2 \alpha \beta x-\alpha \beta=0 \\ & \Rightarrow \quad 3 x^2+2 x-1=0 \\ & \Rightarrow \quad(3 x-1)(x+1)=0 \\ & \Rightarrow \quad x=-1,1 / 3 \\ & \end{aligned}$