If α, β are the zeros of kx^{2} – 2x + 3k such that α + β = αβ then k = ?

We have,

p (x) = x^{2} – 2x + 3k

Now by comparing the given polynomial with ax^{2} + bx + c, we get:

a = 1, b = - 2 and c = 3k

In the question it is given that, are the roots of the given polynomial

∴ =

= -( )

= 2 (i)

:

=

=

= 3k (ii)

Hence, by using (i) and (ii), we have

=

2 = 3k

k =

3