Question

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

Answer 1

Given: x² – 5x + 6 = 0
On comparing this equation with ax² + bx + c = 0, we get:
a = 1, b = –5 and c = 6
α and β are the roots of the equation.
To find the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$, we need to express it in terms of (α + β) and αβ because we know that α + β = –$\frac{b}{a}$ and αβ = $\frac{c}{a}$.
Thus,

$$\begin{gathered} \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta } \\ =\frac{{\alpha }^2+{\beta }^2+2\alpha \beta -2\alpha \beta }{\alpha \beta } \\ =\frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{\alpha \beta } \end{gathered}$$
On substituting the values of α+β = $-\frac{b}{a}$ and αβ = $\frac{c}{a}$ , we get:
$$\begin{gathered} \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\left({\displaystyle \frac{-b}{a}}\right)}^2-2\times \left({\displaystyle \frac{c}{a}}\right)}{{\displaystyle \frac{c}{a}}} \\ =\frac{{\displaystyle \frac{b^2-2ca}{a^2}}}{{\displaystyle \frac{c}{a}}} \\ =\frac{b^2-2ca}{ac} \end{gathered}$$
On substituting a = 1, b = –5 and c = 6, we get:
$$\begin{gathered} \frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\left(-5\right)}^2-2\times 6\times 1}{1\times 6} \\ =\frac{25-12}{6}=\frac{13}{6} \end{gathered}$$