Question

if alpha and beta are zeros of the polynomial x square minus 4 root 3 X + 3 then find the value of alpha + beta minus alpha beta ​

Answer 1

Step-by-step explanation:

${x}^{2} - 4 \sqrt{3} x + 3 = 0 \\ compairing \: equation \: with \: a {x}^{2} + bx + c = 0 \\ a = 1 \: \: \: b = 4 \sqrt{3} \: \: \: c = 3 \\ {b}^{2} - 4ac = {(4 \sqrt{3} )}^{2} - 4 \times 1 \times 3 \\ {b}^{2} - 4ac = 16 \times 3 - 12 \\ {b }^{2} - 4ac = 48 - 12 \\ {b}^{2} - 4ac = 36 \\ using \: formula \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ x = \frac{ - (4 \sqrt{3}) + - \sqrt{36} }{2} \\ x = \frac{ - 4 \sqrt{3} + - 6 }{2} \\ x = \frac{ - 4 \sqrt{3} + 6 }{2} \: \: \: or \: \: \: x = \frac{ - 4 \sqrt{3 } - 6 }{2} \\ x = \frac{ 2( - 2 \sqrt{3} + 3) }{2} \: \: \: or \: \: \: x = \frac{2( - 2 \sqrt{3} - 3)}{2} \\ x = - 2 \sqrt{3} + 3 \: \: \: or \: \: \: x = - 2 \sqrt{3} - 3$

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

$\alpha + \beta - \alpha \beta \\ - 2 \sqrt{3} + 3 + - 2 \sqrt{3} - 3 - ( - 2 \sqrt{3} + 3 \times - 2 \sqrt{3} - 3) \\ = - 4 \sqrt{3} - ( - 2 \sqrt{3} + 3 \times - 2 \sqrt{3} - 3) \\ = - 4 \sqrt{3} - ( { - 2 \sqrt{3}) }^{2} - {3}^{2} \\ = - 4 \sqrt{3} - (4 \times 3) - 9 \\ = - 4 \sqrt{3} - 12 - 9 \\ = - 4 \sqrt{3} - 21$

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