Question

Form the quadratic equations whose zeros 2root3 and -root3

Answer 1

Answer:

$\alpha + \beta = 2 \sqrt{3} - \ \sqrt{3} \\ \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{3} (2 - 1) \\ \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{3}$

And then

$\alpha \beta = 2 \sqrt{3} \times - \sqrt{3} \\ \: \: \: \: \: \: \: = - 6$

So the required quadratic equation is

${x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ = {x}^{2} - \sqrt{3} x + ( - 6) \\ = {x}^{2} - \sqrt{3} x - 6(ans)$

Here is your answer

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