Question

Find a quadratic polynomial whose zeros are
$3 + \sqrt{5 } \div 5 \: and \: 3 - \sqrt{5} \div 5$
​

Answer 1

$Let \: \alpha = \frac{3+\sqrt{5}}{5} \: and$

$\beta = \frac{3-\sqrt{5}}{5} \: are \: zeroes$

$of \: Quadratic \: polynomial$

$i) \alpha + \beta$

$= \frac{3+\sqrt{5}}{5} + \frac{3-\sqrt{5}}{5}$

$= \frac{3+\sqrt{5} + 3-\sqrt{5}}{5}$

$= \frac{ 6}{5} \: --(1)$

$ii) \alpha \beta$

$= \frac{3+\sqrt{5}}{5} \times \frac{3-\sqrt{5}}{5}$

$= \frac{ 3^{2} - (\sqrt{5})^{2}}{25}$

$= \frac{ 9 - 5}{25}$

$= \frac{ 4}{25} \: --(2)$

Therefore.,

$\blue { Form \: of \: a \: Quadratic \: equation }$

$\blue { Whose \: zeroes \: are \: \alpha \: and \: \beta : }$

$= k[ x^{2} - ( \alpha + \beta )x + \alpha \beta ]$

$= k [ x^{2} - \Big( \frac{6}{5}\Big) x + \frac{4}{25} ]$

$If \: k = 25 \: then \: 25x^{2} - 30x + 4$

Therefore.,

$\red{ Required \: polynomial : }$

$\green { = 25x^{2} - 30x + 4}$

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