Question

if alpha and beta are the zeroes of the quadratic polynomial x^2 - 6x + x then find the value of a, if 3alpha + 2beta = 24

Answer 1

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$p(x) = {x}^{2} - 6x + a \\ we \: know \: a {x}^{2} + bx + c \\ \\ a = 1 || b = - 6 || c = a \\ \\ \alpha + \beta(sum \: of \: zeroes) = \frac{ - b}{a} \\ \alpha + \beta = 6 \\ \\ \alpha \beta(product \: of \: zeroes) = \frac{c}{a} \\ \alpha \beta = a \\ \\ = > 3 \alpha + 2 \beta = 24 \\ = > 3 \alpha - \alpha + 2 \beta - \beta = 24 - 6 \\ = > 2 \alpha + \beta = 18 \\ \\ = > 2 \alpha - \alpha + \beta - \beta = 18 - 6 \\ \alpha = 12 \\ \\ \alpha + \beta = 6 \\ \beta = - 6 \\ \\ \alpha \beta = a \\ = > 12 \times ( - 6) = a \\ = > - 72 = a \\ \\ hence \: a = - 72$
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