Question

The equation whose roots are twice the reciprocal of the roots of the quadratic equation x2– px+1=0 is

Answer 1

$Let \: \alpha \:and \: \beta \: are \: two \:roots$

$of \: the \: Quadratic \: equation$

$i) \alpha + \beta = \frac{- Coefficient \: x}{Coefficient \:x^{2}}$

$= \frac{-(-p)}{1}$

$= p \: --(1)$

$ii) \alpha \beta = \frac{Constant \:term}{Coefficient \:x^{2}}$

$= \frac{1}{1}$

$= 1\: --(2)$

$Here, \frac{2}{\alpha} \: and \: \frac{2}{\beta}\:are$

$roots\: of \: a \: Quadratic \: equation.$

$iii) Sum \:of \:the \:roots$

$= \frac{2}{\alpha } + \frac{2}{\beta }$

$= 2\Big( \frac{1}{\alpha } + \frac{1}{\beta} \Big)$

$= 2 \Big( \frac{\beta + \alpha}{\alpha \beta } \Big)$

$= 2 \times \frac{p}{1}\: [ From \: (1) \:and \:(2) ]$

$= 2p \: --(3)$

$iv) Product \:of \:the \:roots$

$= \frac{2}{\alpha } \times \frac{2}{\beta }$

$= \frac{4}{\alpha \beta}$

$= \frac{4}{1}\: [ From \:(2) ]$

$= 4\: --(4)$

$\red{ Required \: Quadratic \: Equation : }$

$\blue{ k[ x^{2} - \Big( \frac{2}{\alpha } + \frac{2}{\beta }\Big) x + \frac{1}{\alpha \beta }]=0 }$

$\implies k( x^{2} - 2px + 4 ) = 0$

$If \: k = 1 \: then \: x^{2} - 2px + 4 = 0$

Therefore.,

$\red{ Required \: Quadratic \: Equation : }$

$\green { \: x^{2} - 2px + 4 = 0 }$

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