Question

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If one Root of the quadratic Polynomial
${2x}^{2} - 3x + p$
is 3, find the other root . Also find the Value of P.​

Answer 1

Hey

$2x {}^{2} - 3x + p = 0 \\ \\ given \: x = 3 \\ \\ = > 2(3) {}^{2} - 3 \times 3 + p = 0 \\ \\ = > 18 - 9 + p = 0 \\ \\ = > p = - 9$

Answer 2

$2 {x}^{2} - 3x + p \: \\ let \: the \: root \: be \: \alpha \: and \: \beta \: \\ and \: it \: is \: given \: that \: \alpha = 3 \: \\ \\ \alpha + \beta = \frac{3}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)\\ \alpha \beta = \frac{p}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2) \\ \\ put \: the \: value \: of \: \alpha \: in \: 1 \\ 3 + \beta = \frac{3}{2} \\ \beta = - \frac{3}{2} \\ now \: put \: value \: of \: \beta \: and \: \alpha \: \: in \: 2 \\ \\ 3 \times \frac{ - 3}{2} = \frac{p}{2} \\ p = - 9$

#Bebrainly

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