Question

find the quadratic polynomial whose zero 2+root3, 2-root3

Answer 1

Heya !!!



Let Alpha = 2+✓3 and Beta = 2-✓3



Sum of zeroes = Alpha + Beta



=> ( 2 + ✓3) + (2-✓3) = 4



And,


Product of zeroes = Alpha × Beta



=> ( 2 + ✓3) ( 2-✓3)




=> (2)² - (✓3)²



=> 4 - 3 = 1



Therefore,



Required Quadratic polynomial = X²-(Sum of zeroes)X + Product of zeroes





=> X²-(4)X + 1




=> X² -4X +1




HOPE IT WILL HELP YOU........ :-)

Answer 2

Hello here
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$let \: \: \alpha = 2 + \sqrt{3} \: \: \: \: \: and \: \: \beta = 2 - \sqrt{3} \\ \\ sum \: of \: zeros = ( \alpha + \beta ) = (2 + \sqrt{3}) + (2 - \sqrt{3} ) = 4 \\ \\ product \: \: of \: zeros = (2 + \sqrt{3} ) \times (2 - \sqrt{3} ) \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = >4 - 3 = 1 \\ \\ so \: the \: \: \: required \: polynomial \: \: is \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ \\ = > {x}^{2} - (4)x + 1 \\ = > {x}^{2} - 4x + 1$
Hence , the required Polynomial is

${x}^{2} - 4x + 1$
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Hope it's helps you.
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