Question

if the sum of the zeroes of the quadratic polynomial ky²+2y-3k is equal to twice then product, find the value of k​

Answer 1

Given

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

Equation- $\sf\ ky^2+2y-3k=0$

To Find :-

We have to find the value of k

Solution :-

Now let us take another quadratic equation

$\sf\ \ ay^2+by+c=0$

On comparing the given equation to this we get :-

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

As we know that

$\sf\ Sum\ of\ zeroes(\alpha+\beta)=\dfrac{-b}{a}\\ \\ \sf\ Product\ of\ zeroes(\alpha\beta)=\dfrac{c}{a}$

We have

$:\implies\sf\ (\alpha+\beta)=\dfrac{-2}{k}\\ \\ \\ :\implies\sf\ \ (\alpha\beta)=\dfrac{-3k}{k}\\ \\ \\ \bullet\sf\ (\alpha+\beta)=2(\alpha\beta)\ \ \ Given\ in\ question\\ \\ \\ :\implies\sf\ \dfrac{-2}{k}= 2\bigg[\dfrac{-3\cancel {k}}{\cancel{k}}\bigg]\\ \\ \\ :\implies\sf\ \dfrac{-2}{k}= 2\times (-3)\\ \\ \\ :\implies\sf\ \ -2= -6k\\ \\ \\:\implies\sf\ \cancel{\dfrac{-2}{6}}=k\\ \\ \\:\implies\underline{\boxed{\sf\purple{ k=\dfrac{1}{3}}}}$

Answer 2

Answer:

1/3

Step-by-step explanation:

If the quadratic equation is of the form ay²+by+c, then

Sum of zeroes of the polynomial = -b/a

Product of zeroes= c/a

So, Sum of zeroes of given polynomial= -2/k

Product of zeroes of polynomial= -3k/k = -3

Now according to question sum is equal to twice the product, so the equation becomes

-2/k = 2*(-3)

⇒ -2 = -6*k

⇒k = -2/-6 = 1/3