Question

If the sum of the zeroes of the quadratic polynomial
$f(t) = kt {}^{2} + 2t + 3k$
is equal to their product. Find
the value of k.
​

Answer 1

REFER TO THE ATTACHMENT ❤️

Answer 2

Let alpha and beta be two zeros of the polynomial.

GIVEN :-

$f(t) = {kt}^{2} + 2t + 3k \\ \\ \alpha + \beta = \alpha \beta$

FORMULA

$\alpha + \beta = \frac{ - coeficient \: of \: {x} }{coeficient \: of \: {x}^{2} } \\ \\ \\ \alpha \beta = \frac{costant}{coeficient \: of \: {x}^{2} }$

SOLUTION :-

$\alpha \beta = \alpha + \beta \\ \\ \implies \frac{3k}{k} = \frac{ - 2}{k} \\ \\ \implies 3k = - 2 \\ \\ \implies \boxed{k = \frac{ - 2}{3} }$