Question

Let alpha and beta be the zeros of the ploynomial x^2-3 then form a quadratic polynomial whose zeros are 1/alpha and 1/beta...

Answer 1

to get a polynomial with roots as reciprocals.. put x = 1/x

1/x²  - 3 = 0
1 -3x² = 0
3x²-1 = 0 is the required equation

Answer 2

$it \: can \: be \: also \: solved \: as \: if \\f(x) = {x}^{2} - 3 \\ then \: the \: polynomial \: with \: roots \: \frac{1}{ \alpha } \: and \: \frac{1}{ \beta } is \\ f( \frac{1}{x} ) = 0 \\ = > {( \frac{1}{x}) }^{2} - 3 = 0 \\ = > \frac{1 - 3 {x}^{2} }{ {x}^{2} } = 0 \\ = > 3 {x}^{2} - 1 = 0$
check the attachment
hope it helps...