Question

If p and q are the zeroes of the polynomial 6y2 - 7y + 2 find a quadratic polynomial whose zeroes are 1/p and 1/q.


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Answer 1

Answer:

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Answer 2

Answer:

$6 {y }^{2} - 7y + 2 \\ 6 {y}^{2} - 4y - 3y + 2 \\ 2y(3y - 2) - 1(3y - 2) \\ (3y - 2)(2y - 1) \\ \\ zeroes \: = 2|3 \: or \: 1 |2 \\ if \: \alpha = 2|3 \: and \: \beta = 1 |2 \\ then \: 1| \: \alpha = 3|2 \: \: and \: 1 | \: \: \beta = 2 \\ \\ now \: the \: quadractic \: polyminial \\ {x}^{2} - (sum \: of \: roots)x + (product \: of \: the \: roots) \\ sum \: of \: the \: roots = 3|2 \: + 2 = 3.5 \\ product \: of \: the \: roots \: 3|2 \: \: \: \: \: \: \: 2 \: = \: \: 3 \\ \\ the \: quadrtic \: is \\ {x}^{2} \: - 3.5x \: + 3$