Question

$if \: one \: of \: the \: zeroes \: of \: the \\ polynomial \: 2x { }^{2} + 3x + \alpha = \\ 1 \div 2 \\ find \: the \: value \: of \: \alpha \: and \: the \\ other \: zero$

Answer 1

For quadratic polynomial
$a {x}^{2} + bx + c$
with zeroes
$\alpha \: and \: \beta .$
Sum of zeroes:


$\alpha + \beta = \frac{ - b}{a} \\ \frac{1}{2} + \beta = \frac{ - 3}{2} \\ \beta = \frac{ - 3}{2} - \frac{1}{2} \\ \beta = - 2$
The other zero is -2.

Now product of zeroes:

$\alpha \beta = \frac{c}{a} \\ \frac{ 1}{2} \times ( - 2) = \frac{c}{2} \\ c = - 2$
The constant term(which you have named alpha in the question) is equal to -2.

Answer 2

Thus, The value of alpha & the 2nd zero is -2.
So, the quadratic equation 2x2 + 3x -2 has two zeroes 1/2 and -2.

Hey mate it will help you out.