Question

form the quadratic equation whose roots are 3+√3 by 2and 3-√3 by 2​

Answer 1

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

$\green{\tt{\therefore{Equation=2x^{2}-6x+3=0}}}$

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

$\green{\underline \bold{Given:}} \\ \tt: \implies Roots\:are =\frac{3+\sqrt{3}}{2}\:and\:\frac{3-\sqrt{3}}{2} \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Quadratic \:equation = ?$

• According to given question :

$\circ\:\tt{let \: \alpha \: and \: \beta \: are \: roots} \\ \tt: \implies sum \: of \: zeroes = \alpha + \beta \\ \\ \tt: \implies sum \: of \: zeroes = \frac{3 + \sqrt{3} }{2} + \frac{3 - \sqrt{3} }{2} \\ \\ \tt: \implies sum \: of \: zeroes = \frac{3 + \sqrt{3} + 3 - \sqrt{3} }{2} \\ \\ \tt: \implies sum \: of \: zeroes = 3 \\ \\ \tt: \implies product \: of \: zeroes = \frac{3 + \sqrt{3} }{2} \times \frac{3 - \sqrt{3} }{2} \\ \\ \tt: \implies product \: of \: zeroes = \frac{ {3}^{2} - (\sqrt{3})^{2} }{4} \\ \\ \tt: \implies product \: of \: zeroes = \frac{9 - 3}{4} \\ \\ \tt: \implies product \: of \: zeroes = \frac{3}{2}$

$\bold{As \: we \: know \: that} \\ \tt: \implies {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes) = 0\\ \\ \tt: \implies {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0 \\ \\ \text{Putting \: given \: values} \\ \tt: \implies {x}^{2} - 3\times x + \frac{3}{2}= 0 \\ \\ \green{\tt: \implies 2{x}^{2} -6x +3 = 0}$