Question

find the zeros of the polynomial x square - 3 and verify the relationship between the zeros and the coefficients

Answer 1

${x}^{2} - 3 = 0 \\ \\ {x}^{2} = 3 \\ \\ x = \pm \sqrt{3} \\ \\ x = \sqrt{3} \: \: and \: \: - \sqrt{3}$

$\text{Let the zeroes be} \ \alpha= \sqrt{3} \ \ and\ \ \beta= - \sqrt{3} \\ \\ coefficients \: are : \\ a = 1 \\ b = 0 \\ c = - 3 \\ \\ \text{Relation between zeroes and coefficient are:} \\ \\ 1.sum \: of \: zeroes: \alpha + \beta = \frac{ - b}{a} \\ \sqrt{3} + ( - \sqrt{3} ) = 0 = \frac{ - (0)}{1} \\ \\ 2.product \: of \: zeroes: \alpha \times \beta = \frac{c}{a} \\ \sqrt{3} \times ( - \sqrt{3} ) = - 3 = \frac{3}{1}$

Answer 2

Here is your solution

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$x {}^{2} - 3 = 0 \\ x {}^{2} = 3 \\ x = + - \sqrt{3}$
$Zeroes\: are:- \\\:α = \sqrt{3} \: and \: β = - \sqrt{3}$
$\\ Coefficient \: are \\ a= 1 \\ b= 0 \\ c= -3 \\ Now \: relationship \: between \: the \: zeros \: and \: the \\ coefficients.\\ \\ 1.sum \: of \: zeroes \\ α+β= \frac{ - b}{a} \\ \\ \sqrt{3} + ( - \sqrt{ 3} ) = \frac{ - (0)}{1} \\ \\ 0 = 0$

$2.product \: of \: zeroes \\ α× \: β = \frac{c}{a} \\ \\ \sqrt{3} \times - \sqrt{3} = \frac{ - 3}{1} \\ - 3 = -3$

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