Question

If the sum of the zeros of the quadratic polynomial kx square+2x-3k is equal to twice their product, find values of k

Answer 1

sum = 2(product)

-2/k=2(-3k/k)

k=3

Answer 2

$\mathfrak{\huge{Answer:}}$

$\mathbb{GIVEN}$

The equation : $\tt{kx^{2} + 2x - 3k}$

The sum of the zeroes = 2 ( product of the zeroes )

$\mathbb{TO\:FIND}$

The value of k in the given equation

$\mathbb{METHOD}$

We know that when $\sf{\alpha}$ and $\sf{\beta}$ are the two zeroes of any polynomial p(x), then :-

Sum of the zeroes = $\sf{\alpha + \beta = \frac{-b}{a}}$

Product of the zeroes = $\sf{\alpha\beta = \frac{c}{a}}$

If we write the given information ( The sum of the zeroes = 2 ( product of the zeroes )) in the algebraic form, we get that :-

$\sf{\alpha + \beta = 2 ( \alpha \beta)}$

In other words, we mean to say that :

$\sf{\frac{-b}{a} = 2(\frac{c}{a})}$

Here, we're given the values of the variables as :

a = k

b = 2

c = (-3k)

Bring the values in the equation and solve the formed equation:

$\sf{\frac{-2}{k} = 2(\frac{(-3k)}{k}}$

=》 $\sf{(-2) = 2(-3k)}$

=》 $\sf{k = \frac{1}{3}}$

Thus, the value of k will be $\tt{\huge{\frac{1}{3}}}$