Question

If alpha and Beta are the zeroes of the polynomial 15x²-x-4 then find a quadratic polynomial whose zeroes are 2 alpha and 2beta

Answer 1

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Answer:-

{x}^{2} - 2x - 16[/tex

Step-by-step explanation:

$Quadratic \: polynomial \: is \: {15x}^{2} - x - 4 \\ Given \: a \: quadratic \: polynomial \: is \: a {x}^{2} + bx + c \\ sum \: of \: the \: zeroes \: is \: \alpha \: and \beta \: is \ - \frac{a}{b} \\ and \: Product \: of \: the \: zeroes \: is \: \frac{c}{b} further. \\ If \: sum \: of \: s \: and \: product \: of \: zeroes \: is \: p \: \\ then, \: queadratic \: polynomial \: is \: {x}^{2} - sx + p.\\ so,\: {x}^{2} - sx + p = {x}^{2} - ( \alpha + \beta )x + \alpha \times \beta \\ <u>Hence</u> ,\: for \: 15 {x}^{2} - x - 4.\\ \: \: \: \: \: \: \: \: \: \alpha + \beta = \<strong>frac{1}{15}</strong> \\ and \: \alpha \times \beta = <strong> - \frac{4}{15}</strong> \\ \\ We \: now \: desire \: \: quadratic \: polynomial \: \\ where \: zeroes \: are \: 2 \alpha \: and \: 2 \beta. \\ \\ As \: sum \: of \: root \: is \: \\ = > 2 \alpha + 2 \beta = 2( \alpha + \beta ) = 2 \times \frac{1}{15} = <strong>\frac{2}{15</strong>} \\ And \: product \: of \: root \\ = > 2 \alpha \times 2 \beta = 4 \alpha \beta = 4 \times \frac{ - 4}{15} = <strong> \frac{ -</strong> <strong>16}{15}</strong> \\ And \: quadratic \: polynomial \: is \: <strong>{x}^{2} - \frac{2}{15} x - \frac{16}{15} </strong> \\ \\ As \: zeroes \: are \: not \: affected \: by \: multiplying \\ each \: term \: of \: polynomial \: by \: a \: consant. \\ \\ We \: can \: say \: quadratic \: polynomial \: \\ = > \: <strong> {x}^{2} - 2x - 16$

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