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2 November, 08:20

if alpha and beta are the zeroes of the polynomial 6y^2 (six y squared) - 7y + 2, find a quadratic polynomial whose zeroes are 1/alpha + 1/beta

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  1. 2 November, 09:38
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    6y² - 7y + 2 = 0

    y=7±√ (49-4*6*2) / 12

    = 7±1 / 12

    = ½ or ⅔

    so we want our polynomial to have roots of 2 and 3/2

    (x-2) (x-3/2) = 0

    x² - 7/2x + 3 = 0
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