Question

If alpha and beta are the zeros of the quadratic polynomial 3x^2-4x-7, find a quadratic polynomial whose zeros are 1/alpha and 1/beta

Answer 1

3x^2 -4x -7

alpha beta are zeroes of it so

$\alpha + \beta = \frac{ - ( - 4)}{3} = \frac{4}{3} \\ \alpha \beta = \frac{7}{3}$

if roots are

$\frac{1}{ \alpha } \: \: and \: \: \frac{1}{ \beta }$

so equation is

$(x - \frac{1}{ \alpha } )(x - \frac{1}{ \beta } ) = 0 \\ \\ {x}^{2} - x( \frac{1}{ \alpha } + \frac{1}{ \beta } ) + \frac{1}{ \alpha \beta } = 0 \\ \\ {x}^{2} - x( \frac{ \beta + \alpha }{ \alpha \beta } ) + \frac{1}{ \alpha \beta } = 0$

putting values in equation

${x}^{2} - x( \frac{ \frac{4}{3} }{ \frac{7}{3} } ) + \frac{1}{ \frac{7}{3} } = 0 \\ \\ {x}^{2} - \frac{4x}{7} + \frac{3}{7} = 0 \\ \\ 7 {x}^{2} - 4x + 3 = 0$