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Question

If sum of the zeroe of the polynomial Ky^2+2y-3K is twice their product of zeros then find the value of K

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Solution

Quadratic equation is

Ky² + 2y - 3k

Let α and β are its two zeros

The sum of zeros = α + β = -b/a
product of zeros = αβ =ab
Given condition. α + β = 2× αβ
-2 / k = -2×3k/k
-6k = -2
k =2/6
k = 1/3

hope you got it.

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