Question

form a quadatic equation whose sum and product of zeros are zeros are ⅔ and -⅓

Answer 1

$sum \: \: of \: \: zeroes = \frac{2}{3}$

$pdt \: \: of \: \: zeroes = \frac{ - 1}{3}$

$so \: \: the \: \: equation \: \: is$

$x^{2} - \frac{2}{3} x + ( \frac{ - 1}{3} )$

$x ^{2} - \frac{2}{3} x - \frac{1}{3}$

$\frac{3x ^{2} - 2x - 1}{3}$

$\frac{1}{3} (3x ^{2} - 2x - 1)$

$3x ^{2} - 2x - 1 \: \: is \: \: the \: \: required \: \: equation$

Answer 2

Answer:

root are 2/3 and -1/3

so let

sum of roots=2/3

and product of roots =-1/3

on substituting values in

k(x^2- sx +p)

here s = sum of roots

and p= product of roots

so,

k(x^2-2/3x-1/3)

=k(3x^2/3-2/3x-1/3)

LCM=3

so value of k =3

so quadratic equation will be

3x^2-2x-1