Question

If α and β are the zeros of the polynomials 3y²-8 y + 4, find a quadratic polynomial whose roots are 1/α and 1/β

Answer 1

I think

(3·(1/y)^2 - 8·(1/y) + 4)·y^2

= (3/y^2 - 8/y + 4)·y^2

= 3 - 8y + 4y^2

= 4y^2 - 8y + 3

Answer 2

4y² - 8y + 3 = 0

Explanation:

given 3y² - 8y + 4 ← in standard form

with a = 3, b = - 8, c = 4

and that α and β are roots of the polynomial, then

α + β = -
`(b)/(a)` =
`(8)/(3)`, and

αβ =
`(c)/(a)` =
`(4)/(3)`

sum of new roots =
`(1)/(\alpha )` +
`(1)/(\beta )`

=
`(\alpha +\beta )/(\alpha \beta )`

=
`((8)/(3) )/((4)/(3) )` = 2

product of new roots =
`(1)/(\alpha )` ×
`(1)/(\beta )`

=
`(1)/(\alpha \beta )` =
`(1)/((4)/(3) )` =
`(3)/(4)`

Hence the required equation is

y² - 2y +
`(3)/(4)` = 0 ( multiply through by 4 )

4y² - 8y + 3 = 0