Question

If alpha and beta are roots of the equation 2x^2-(p+1)x+(p-1)=0 and alpha-beta =alpha×beta,then the value of p is a) 3 b)-2 c)1 d)2

Answer 1

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$If \: \alpha \:and\: \beta\: are \:roots\: of\: the\: equation\\ 2x^2-(p+1)x+(p-1)=0 \:and\\ \alpha - \beta =\alpha \beta, \:then \:the \:value\: of\: p \:is:$$<ol type ="a"></p><p><li>3</p><p><li>-2<li>1<li>2</ol>$

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$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$

$\boxed{\red{\bold{Sum \:of \:zeroes:}}}$

$\purple{\implies \alpha + \beta = \displaystyle{\frac{-b}{a}}}$

$\purple{\implies} \alpha + \beta = \displaystyle{\frac{-[-(p+1)]}{2}}$

$\purple{\implies} \alpha + \beta = \displaystyle{\frac{p+1}{2}}$

$\boxed{\red{\bold{Product \: of \: zeroes:}}}$

$\purple{\implies \alpha \beta = \displaystyle{\frac{c}{a}}}$

$\purple{\implies}\alpha \beta = \displaystyle{\frac{p-1}{2}}$

$\boxed{\red{\bold{Difference \: of\: zeroes:}}}$

$\blue{\implies \alpha - \beta = \displaystyle \sqrt{(\alpha + \beta)^2 - 4 \alpha \beta}}$

$\blue{\implies}\alpha - \beta = \displaystyle \sqrt{\left( \displaystyle{\frac{p+1}{2}} \right)^2 - 4 × \frac{p-1}{2}}$

$\blue{\implies}\alpha - \beta =\displaystyle \sqrt{\displaystyle{\frac{p^2 +2p +1}{4}} - \frac{4p-4}{2}}$

$\blue{\implies}\alpha - \beta = \displaystyle \sqrt{\displaystyle{\frac{p^2 + 2p + 1 - 8p + 8}{4}}}$

$\blue{\implies}\alpha - \beta = \displaystyle \sqrt{\displaystyle{\frac{p^2 - 6p+9}{4}}}$

$\blue{\implies}\alpha - \beta = \pm \displaystyle{\frac{p-3}{2}}$

$\boxed{\red{\bold{Given\: relation:}}}$

$\green{\alpha - \beta = \alpha \beta}$

$\green{\boxed{Putting\: \alpha - \beta = +\left( \displaystyle{\frac{p-3}{2}} \right)}}$

$\green{\implies}\displaystyle{\frac{p-3}{2}} = \frac{p-1}{2}$

$\green{\implies}It\: is\; not \:possible$

$\green{\boxed{So \:taking\: \alpha - \beta = \displaystyle{\frac{-\left( p-3 \right)}{2}}}}$

$\green{\implies}\displaystyle{\frac{-p+3}{2}} = \frac{p-1}{2}$

$\green{\implies}-p+3 = p-1$

$\green{\implies}2p = 4$

$\green{\boxed{\bold{p=2}}}$

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