Question

Find a quadratic polynomial with the given numbers as the sum and product of its

zeroes respectively 2 + √ and 2 - √​

Answer 1

$let \: the \: \alpha \: and \: \beta \: be \: zeros \: of \\ polynomial$

$Now, \\ let \: \alpha = \: 2 + \sqrt{3} \:,\: \beta=2- \sqrt{3}$

$we \: know \: that,\: \\ sum \: of \: zero = \alpha + \beta \: and \\ product \: of \: zero = \alpha \times \beta$

$so, \: \: \alpha + \beta =2 + \sqrt{3}\:+2 - \sqrt{3} \: \\ \: \implies\alpha + \beta = (2 + 2) = 4$

$\alpha \times \beta = (2 + \sqrt{3})\times (2-\sqrt{3}) \\ \alpha \times \beta = 4 + 2 \sqrt{3} - 2 \sqrt{3} - 3 \\\alpha \times \beta = (4 - 3) = ( \: 1)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$we \: know \: that \: for \: any \: quadratic \\ euation \: the \: sum \: of zero \: = \frac{ - b}{a} \\ and \: product \: of \: \: zero = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \:$

$so,sum \: of \: zero = \frac{ - b}{a} = \frac{4}{1} \\ sum \: of \: product = \frac{c}{a} = \frac{1}{1}$

$\therefore \: a = 1 \:, \: b = - 4 \:, \: c = 1$

$qadratic \: is \: of \: the \: form \: \\ ax^{2} + bx + c = 0$

$1x^{2} - 4x+1 \: is \: the \: quadratic \: \\ eqation$

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