Question

find the zeroes of the quadratic polynomials and verify the relationship between the zeroes and the coefficients:​

Answer 1

Answer:

$5 {y}^{2} + 10y \: = 0 \\ 5y(y + 2) = 0 \\ y = 0 \: or \:y = - 2$

Answer 2

Answer:-

5y² + 10y = 0

5y (y + 2) = 0

y + 2 = 0/5y

y + 2 = 0

y = (-2).

Relationship between zeroes and coefficient of a quadratic equation:-

Let the quadratic equation be ax² + bx + c = 0 where a,b,c belonging real numbers and a ≠ 0.

$its \: roots \: are \: \alpha \: and \: \beta \: where \: \alpha = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} \: and \: \\ \\ \beta \: = \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ sum \: of \: roots \: (zeroes) \: = \frac{ - b + \sqrt{ {b}^{2} - 4ac} }{2a} + \frac{ - b - \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ = \frac{ - b + \sqrt{ {b}^{2} - 4ac} - b - \sqrt{ {b}^{2} - 4ac } }{2a } \\ \\ = \: \frac{ - 2b}{2a} \\ \\ \alpha + \beta = - \frac{b}{a} \\ \\ product \: of \: the \: roots \: = \alpha \beta \\ \\ = \: { \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} } \times { \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a} }\\ \\ = \frac{ {( - b)}^{2} - ( \sqrt{ {b}^{2} - 4ac })^{2} }{ {(2a)}^{2} } \\ \\ = \frac{4ac}{4 {a}^{2} } \\ \\ \alpha \beta = \frac{c}{a}$ .

Thus , the sum and product of roots show the relationship between zeroes and coefficients.