Question

Find quedratic polynomial if sum of product of zeros are 2√3and-√3

Answer 1

Step-by-step explanation:

Hope it is useful ........

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Quadratic\:eqn\:\:x^{2}-2\sqrt{3}x-\sqrt{3}=0}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Sum \: of \: zeroes = 2 \sqrt{3} \\ \\ \tt: \implies Product \: of \: zeroes = - \sqrt{3} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Quadratic \: polynomial = ?$

• According to given question :

$\tt{Let \: zeroes \: be \: \alpha \: and \: \beta } \\ \\ \tt \circ \: \alpha + \beta =2 \sqrt{3} \\ \\ \tt \circ \: \alpha \times \beta = - \sqrt{3} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies { x }^{2} - (Sum \: of \: zeores)x + (Product \: of \: zeroes) = 0\\ \\ \tt: \implies {x}^{2} - ( \alpha + \beta )x + ( \alpha \beta ) = 0 \\ \\ \text{Putting \: given \: value} \\ \tt: \implies {x}^{2} - 2 \sqrt{3} x + ( - \sqrt{3} ) = 0 \\ \\ \tt: \implies {x}^{2} - 2 \sqrt{3} x - \sqrt{3} = 0 \\ \\ \green{\tt\therefore quadratic \: eqn = {x}^{2} - 2 \sqrt{3} x - \sqrt{3} = 0} \\ \\ \blue{\bold{Some \: related \: formula}} \\ \orange{ \tt \circ \: D = {b}^{2} - 4ac} \\ \\ \orange{ \tt \circ \: x = \frac{ - b \pm \sqrt{D} }{2a} }$