Question

Find a quadratic polynomial each with the given numbers as the zeros of the polynomial
2√3,-2√3​

Answer 1

please see the attached photo for your answer

Answer 2

$\small \bf \pink{hey \: mate \: here \: is \: ur \: ans} \\ \\ \small \bf \pink{ \huge \: solution} \\ \\ \small \bf \pink{given \: that} \\ \\ \small \bf \pink{2 \sqrt{3} \: \: and \: - 2 \sqrt{3} } \\ \\ \small \bf \pink{ \underline{ \underline{step \: by \: step \: explaination}}} \\ \\ \small \bf \pink{let \: be \: } \\ \\ \small \bf \pink{ \alpha = 2 \sqrt{3} } \\ \\ \small \bf \pink{ \beta = - 2 \sqrt{3} } \\ \\ \small \bf \pink{ \huge \: now} \\ \\ \small \bf \pink{ \underline{ \underline{using \: this \: formula}}} \\ \\\small \bf \pink{ (x - \alpha )(x - \beta )} \\ \\ \small \bf \pink{(x - 2 \sqrt{3} )(x + 2 \sqrt{3} )} \\ \\ \small \bf \pink{ {x}^{2} + 2 \sqrt{3}x - 2 \sqrt{3}x - 12 = 0 } \\ \\ \small \bf \pink{ {x}^{2} - 12 \: ans}$