Question

་་་་་་find the quadratic polynomial whose zeros are 7 + root 3 and 7 minus root 3

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Answer 1

Question :-

Find the quadratic polynomial whose zeros are 7 +√3 and 7 - √3 .

Answer :-

$\boxed{\sf{ {x}^{2} - 14x + 46 = 0}} \:$

To Find :-

→ Quadratic polynomial .

Explanation :-

$\sf{let \: given \: zeros \: are \: \alpha \: \: and \: \beta } \\ \\ \sf{sum \: of \: zeros \: = \alpha + \beta } \\ \\ \implies \sf{ \alpha + \beta = 7 + \sqrt{3} + 7 - \sqrt{3} } \\ \\ \implies \: \boxed{ \alpha + \beta = 14} \\ \\ \sf{ now \: product \: of \: zeros \: = \alpha \beta } \\ \\ \implies \: \alpha \beta = (7 + \sqrt{3} )(7 - \sqrt{3} ) \\ \\ \implies \: \alpha \beta = 49 - 3 \\ \\ \implies \: \boxed{ \alpha \beta = 46}$

Then required quadratic polynomial is ,

$\implies \sf{ {x}^{2} - ( \alpha + \beta )x + \alpha \beta = 0} \\ \\ \implies \boxed{\sf{ {x}^{2} - 14x + 46 = 0}}$