Question

find the zeroes of the polynomial x²-3and verify the relationship between the zeros and coeffients​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$p(x) = {x}^{2} - 3 = 0\\ {x}^{2} = 3 \\ x = \sqrt{3} \\ x = + \sqrt{3} \:a nd \: x = - \sqrt{3} \\ verification \\ let \: \alpha \: and \: \beta \: be \: the \: zeroes \: of \: p(x) \\ here \: a = 1 \\ \: \: \: \: \: \: \: \: \: \: b = 0 \\ \: \: \: \: \: \: \: \: \: \: c = 3 \\ sum \: of \: zeroes = \alpha + \beta = \frac{ - b}{a} \\ = \frac{ - 0}{1} = 0 \\ product \: of \: zeroes = \alpha \beta \\ = \frac{c}{a} \\ = \frac{3}{1} \\ = 3$

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