\(x^2-22x+105=0\) সমীকরনের বীজদ্বয় \(\alpha, \beta\) হলে \(\cfrac{1}{\alpha}+\cfrac{1}{\beta}\) এর মান নির্ণয় করো ।
Madhyamik 2024
\(α+β=22\) এবং \(αβ=105 \)
\(\cfrac{1}{\alpha}+\cfrac{1}{\beta}\)
\(=\cfrac{\beta+\alpha}{\alpha\beta}\)
\(=\cfrac{22}{105}\)
\(∴\cfrac{1}{\alpha}+\cfrac{1}{\beta} = \cfrac{22}{105} \) (Answer)