Question

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Answer 1

Answer:

sum of zero=1+-3=-2

product of zeros=1*-3=-3

polynomial=x²-(sum)x+product=x²+2x-3

verify

sum of zero=-(coefficient of x)/coefficient of x²

-2=-2/1

product of zeros=constant/coefficient of x ²

-3=-3

verified

Answer 2

$Let \: the \: zeroes \: of \: the \: polynomial \: be \: \alpha \: and \: \beta \: respectively. \\ \\ \\ Quadratic \: Polynomial = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ \\ Given , \\ \alpha = 1 \\ \\ \beta = - 3 \\ \\ Quadratic \: Polynomial = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ Quadratic \: Polynomial = {x}^{2} - ( 1 + ( - 3) )x + ( 1 \times - 3) \\ Quadratic \: Polynomial = {x}^{2} - (1 - 3)x + ( - 3) \\ Quadratic \: Polynomial = {x}^{2} - ( - 2 )x - 3 \\ Quadratic \: Polynomial = {x}^{2} + 2 x - 3 \\ \\ \\ \\ VERIFICATION \: of \: zeroes \\ \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha + \beta = 1 + ( - 3) \\ \alpha + \beta = 1 - 3 \\ \alpha + \beta = - 2 \\ \\ \frac{ - b}{a} = \frac{ - 2}{1} \\ \frac{ - b}{a} = - 2 \\ \\ Therefore \: it \: is \: verified \: that \: \alpha + \beta = \frac{ - b}{a} \\ \\ \\ \alpha \beta = \frac{c}{a} \\ \\ \alpha \beta = 1 \times - 3 \\ \alpha \beta = - 3 \\ \\ \frac{c}{a} = \frac{ - 3}{1} \\ \frac{c}{a} = - 3 \\ \\ Therefore \: it \: is \: verified \: that \:\alpha \beta = \frac{c}{a} .$

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