Question

if the sum of the zeroes of p(x) =
$kt { \: }^{2} + 2t + 3k$
is equal to their product . then find k

Answer 1

we know that sum of zeros = -b/a
and product of zeros = c/a
so according to question
$\frac{ - b}{a} = \frac{c}{a} \\ \frac{ - 2}{k} = \frac{3k}{k} \\ \frac{ - 2}{k} = 3 \\ k = \frac{ - 2}{3}$

Answer 2

We know that ,
in quadratic equation ,
ax² +bx +ç

Sum of zeroes = -b/a

product of zeroes =ç/a

Similarly in kt² + 2t +3k

sum of zeroes = -2/k

product of zeroes = 3k/k= 3

Now Given,

Sum of zeroes= product of zeroes
→-2/k = 3

→k/(-2)= 1/3

→k= -2/3