Question

If one of the zeros of the polynomial 3y^2+13y-p is the reciprocal of the other one then find p

Answer 1

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$\sf{If \: one \: of \: the \: zeros \: of \: the \: polynomial \: 3y^2+13y-p \: } \\ \sf{ is \: the \: reciprocal \: of \: the \: other \: one \: then \: find \: p}</p><p>$

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$\sf{if \: \alpha \: is \: one \: of \: the \: zeroes \: then \: other \: is \: \frac{1}{ \alpha } }$

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$\sf{p \: where \: p \: is \: constant(c)}$

$\sf{ \to \: let \: p(y) = 3 {y}^{2} + 13y - p} \\ \\ \sf{\to \: we \: know..} \\ \sf{if \: f(x) = a {x}^{2} + bx + c} \\ \sf{ \boxed{ \red{sum \: of \: zeroes = \frac{ - b}{a} }}} \\ \sf{ \boxed{ \red{ product \: of \: zeroes = \frac{c}{a} }}} \\ \\ \sf{ \to \: by \: comparing \: p(y) \: with \: f(x)} \\ \\ \alpha + \frac{1}{ \alpha } = \frac{ - 13}{3} ( \alpha \: and \: \frac{1}{ \alpha } are \: zeroes \: of \: p(y) \\ \green{ \sf{ \to \: we \: dont \: need \: this}} \\ \\ \alpha \times \frac{1}{ \alpha } = \sf{ \frac{ - p}{3} } \\ \\ \sf{ = > 1 = \frac{ - p}{3} } \\ \\ \sf { \purple{ \boxed{= > p = \frac{ - 1}{3} }}} \\ \\ \sf{ \bold{ \green{\% \: required \: answer...}}}$

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Answer 2

Answer:

Hey

Here is your answer,

5z^2 + 13z - p = 0

Let the zeroes be alpha and 1/ alpha

Product of zeroes = c/a

Alpha x 1/alpha = -p/5

-p/5 = 1

p = -5