If the sum of the zeroes of the quadratic polynomial $f(t)\ =\ kt^2\ +\ 2t\ +\ 3k$ is equal to their product, then find the value of $k$.

Given:

The sum of the zeroes of the quadratic polynomial $f(t)\ =\ kt^2\ +\ 2t\ +\ 3k$  is equal to their product. 

To do:

Here, we have to find the value of $k$.

Solution:  

We know that, 

The standard form of a quadratic polynomial is $at^2+bt+c$, where t is the variable, a, b and c are constants and $a≠0$.

Comparing the given polynomial with the standard form of a quadratic polynomial, 

$a=k$, $b=2$ and $c=3k$

Sum of the zeros$=\frac{-b}{a}=\frac{-2}{k}$.

Product of the zeros$=\frac{c}{a}=\frac{3k}{k}=3$.

Therefore, 

$\frac{-2}{k}=3$

$k=\frac{-2}{3}$

The value of $k$ is $\frac{-2}{3}$.

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Updated on: 10-Oct-2022

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