Question

the zeroes are 2/5 and 21 form quadratic equation​

Answer 1

Step-by-step explanation:

we know that

$\alpha = \frac{2}{5} \\ \beta = 21 \\ \alpha + \beta = \frac{107}{5} \\ \alpha \beta = \frac{42}{5} \\ \\ k( {x}^{2} - ( \frac{107}{5} )x + \frac{42}{5} ) = 0 \\ \\ \frac{k}{5} (5 {x}^{2} - 107x + 42) = 0\\ \\ 5 {x}^{2} - 107x + 42 = 0$