Question

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

Answer 1

Hello !
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Here , 

Quadratic equation is 

 Ky² + 2y - 3 

Let α and β are its two zeros 
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The sum of zeros = α + β = -b/a 
product of zeros   = αβ     = c/a 

An it is Given that=> -b/a = 2 ( c/a) 
                                  
-2 / k    = -6/ k 
-6k       = -2k 
6k - 2k = 0 
4k        = 0
k          = 0/4 
k          = 0 
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But if K = 0 then y² term will be vanished and then it will no more a quadratic equation 
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If Quadratic polynomial have real zeros then , 
Discriminant   ≥ 0
b² - 4ac           ≥ 0
4 - 4 x -3 x k   ≥ 0
4 +12k            ≥ 0 
k                     ≥ -4/12 
k                     ≥ -1/3

Answer 2

Let a and b are the zeros of quadratic equation ; Ky² + 2y -3 = 0
sum of zeros = -( coefficient of x)/coefficient of x²) = -(2)/K = -2/K
product of zeros = constant/coefficient of x²
= -3/K

A/C to question,
sum of zeros = 2 × product of zeros
-2/K = 2(-3)/K
-2 =-6. but -2≠ -6
here you can see that
LHS ≠ RHS

this question is given wrong data , it's not possible .

we can't find out K value by this condition .
but qudartic polynomial have real zeros then
Discrimant ≥ 0
( 2)² - 4(-3)K ≥ 0
1 + 3K ≥ 0
K ≥ -1/3
hence, for defining quadratic polynomial , K must be greater then -1/3