Question

If the sum of the zeros of a quadratic polynomial ky² + 2y - 3k is equal to twice their product, find the value of K ?​

Answer 1

Given a polynomial

$\sf\:\: ky{}^{2}+2y-3k$

$\sf \:\:Sum\:of\: zero's =\dfrac{-b}{a}\\ \\ \sf\:\:\: Product\:of\:zeros=\dfrac{c}{a}$

$\sf\bullet a= k\\ \\ \sf\bullet b=2\\ \\ \sf\bullet c= -3k$

Now the sum of zero's

$\sf\implies Sum\:of\: zero's=\dfrac{-b}{a}\\ \\ \sf\implies {\blue{sum\:=\dfrac{-2}{k}}}$

$\sf\implies Product=\dfrac{c}{a}\\ \\ \sf\implies \dfrac{(-3\cancel{k})}{\cancel{k}}\\ \\ \sf\implies {\purple{Product=(-3)}}$

Now it is given that

$\sf\bullet Sum\:of\: zero's=2\times product\\ \\ \sf\implies \dfrac{-2}{k}=2\times (-3)\\ \\ \sf\implies -2=k\times -6\\ \\ \sf\implies \cancel{\dfrac{-2}{-6}}=k\\ \\ \sf\implies k=\dfrac{1}{3}$

Answer 2

Answer:

⋆ Given Polynomial : ky² + 2y – 3k

Here : a = k,⠀b = 2,⠀c = – 3k

$\underline{\bigstar\:\textsf{According to the Question :}}$

$\dashrightarrow\texttt{\:\:Sum of Zeroes = 2(Product of Zeroes)}\\\\\\\dashrightarrow\tt\:\: \alpha + \beta = 2( \alpha \beta)\\\\\\\dashrightarrow\tt\:\: \dfrac{ - \:b}{a} =2 \times \dfrac{c}{a}\\\\\\\dashrightarrow\tt\:\: \dfrac{ - \:2}{k} =2 \times \dfrac{ - \:3k}{k}\\\\\\\dashrightarrow\tt\:\: - \:2 =2 \times - \:3k\\\\\\\dashrightarrow\tt\:\: - \:2 = - \:6k\\\\\\\dashrightarrow\tt\:\: \dfrac{ - \:2}{ - \:6} =k\\\\\\\dashrightarrow\:\: \large\underline{\boxed{\red{\tt k = \dfrac{1}{3}}}}$