Question

the quadratic polynomial whose zeroes are √2 and √6/2 are​

Answer 1

Step-by-step explanation:

$Let, \: \alpha \: and \: \beta \: be \: the \: zeroes \: of \: polynomial$

$\color{blue}\boxed{ \huge{ \mathbb{ \colorbox{pink}{Given :-}}}}$

$\bold{ \: : \implies\alpha = \sqrt{2} }$

$\bold{ \: : \implies\beta = \frac{ \sqrt{6} }{2} }$

$\mathtt{We \: know \: that,}$

$sum \: of \: zeroes = \alpha + \beta = \sqrt{2} + \frac{ \sqrt{6} }{2}$

$\implies \frac{2 \sqrt{2} + 6}{2}$

$\implies \frac{ \not2( \sqrt{2} + 3 )}{ \not2}$

$product \: of \: zeroes = \alpha \times \beta = \sqrt{2} \times \frac{ \sqrt{6} }{2}$

$\implies \frac{ \sqrt{12} }{2}$

$\mathtt{We \: also \: know \: that,}$

${x}^{2} - ( \alpha + \beta )x + ( \alpha \times \beta ) = 0$

${x}^{2} - \sqrt{2} x + 3x + \frac{ \sqrt{12} }{2} = 0$

$2 {x}^{2} - (2 \sqrt{2} x + 6x) + \sqrt{12} = 0$

$2 {x}^{2} - (2 \sqrt{2} x + 6x) + \sqrt{12} = 0 \: is \: the \: required \: polynimal$