Let \(\alpha = \frac 23\) and \(\beta = \frac{-1}4\)
Sum of the zeroes = (α + β)
\(= \frac{2}{3} + (\frac{-1}4 )\)
\(=\frac 5 {12}\)
Product of the zeroes = αβ
\(= \frac{2}{3} \times( \frac{-1}4 )\)
\(= \frac{-1}6\)
Required quadratic polynomial is
\(x^2 - (\alpha + \beta )x + \alpha \beta = x^2 - (\frac 5{12})x - (\frac{-1}6)\)
\(= \frac 1{12} (12x^2 - 5x - 2)\)
Sum of the zeroes = \(\frac 5{12} = \frac{\text{(-coefficient of x)}}{\text{coefficient of }x^2}\)
Product of zeroes = \( \frac{-1}6 = \frac{\text{constant term}}{\text{coefficient of }x^2}\)
