Question

Find a quadratic polynomial whose zeros are 3+root5/2 and 3-root5/2

Answer 1

Question

Find a quadratic polynomial whose zeros are 3+root5/2 and 3-root5/2

Answer

here we know that

$sum \: of \: zeros \: = \alpha + \beta \\ \\ product \: of \: zeros \: = \alpha \beta \: \\ now \: let \: 3 + \sqrt{ \frac{5}{2} } \: be \: \alpha \: and \: 3 - \sqrt{ \frac{5}{2} } be \: \beta \\ \\ then \: sum \: of \: zeros \: = 3 + \sqrt{ \frac{5}{2} } + 3 - \sqrt{ \frac{5}{2} } \\ \\ so sum \: of \: zeros = \: 3 + 3 \: \: \: (coz \: - \sqrt { \frac{5}{2} } + \sqrt{ \frac{5}{2} } will \: cancel \: out \: ) \\ \\ sum \: of \: zero \: be \: = 6 \\ \\ now \: product \: of \: zeros \: = (3 + \sqrt{ \frac{5}{2} }) ( 3 - \sqrt{ \frac{5}{2} } ) \: \: \: {by \: using \: property \: of \: {a}^{2} - {b}^{2} } \\ \\ product \: of \: zeros = {3}^{2} - { \sqrt{ \frac{5}{2} } }^{2} \\ product \:of \: zeros = 9 - \frac{5}{2} \\ product \: of \: zero \: = \frac{18 - 5}{2} \\ product \: of \: zeros = \frac{13}{2} \\ \\ by \: using \: formula \: = {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ \\ as \: we \: know \: \alpha + \beta = sum \: of \: zeros \: = 6 \\ so \: \alpha + \beta = 6 \\ also \: \alpha \beta = product \: of \: zeros \: = \frac{13}{2} \\ so \ \alpha \beta = \frac{13}{2} \\ \\ by \: putting \: value \: in \: \: formula \: \\ \\ {x}^{2} - 6x + \frac{13}{2} \\ \\ \\ hence \: required \: polynomial \: is \: {x}^{2} - 6x + \frac{13}{2}$

hope it helps

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