Since α and β are the zeroes of the quadratic polynomial f(x) = 6x2 + x − 2,
Sum of the zeroes = (α + β)
\(=\frac{-b}a \)
\(=\frac{-1}6\)
The product of the zeroes = αβ
\(=\frac ca \)
\(= \frac{-2}6\)
\( = \frac {-1}3\)
Now,
\(\frac \alpha\beta+ \frac \beta \alpha = \frac{(\alpha^2 + \beta^2)}{\alpha \beta}\) (by taking LCM)
∵ (α + β)2 = α2 + β2 + 2αβ
\(\frac \alpha\beta+ \frac \beta \alpha = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta}\)
By substitution the values of the sum of zeroes and products of the zeroes, we will get
\(= \cfrac{\left(\frac{-1}6\right)^2 - 2\left(\frac {-1}3\right)}{\left(\frac {-1}3\right)}\)
\(= \cfrac{\left(\frac 1{36}+ \frac 23\right)}{\left(\frac{-1}3\right)}\)
\(= \cfrac{\left(\frac {1 + 24}{36}\right)}{\left(\frac{-1}3\right)}\)
\(= \cfrac{\left(\frac { 25}{36}\right)}{\left(\frac{-1}3\right)}\)
\(= \frac{-25}{12}\)
\(\frac \alpha \beta + \frac \beta \alpha= \frac{-25}{12}\)
