Question

If the sum of zeroes of the quadratic polynomial KY square+2y-3k is equal to twice their product find the value of k ??

Answer 1

$\:\: \underline{\underline{\bf{\large\mathfrak{~~Solution~~}}}}$

the given equation is =ky²+2y-3k

the sum of the zeroes of the given equation is

=-2

and the product of the zeroes is

=-3k

now according to the Question...

$= > - 2 = 2( - 3k) \\ = > k = \frac{1}{3}$

$:\: \underline{\underline{\bf{\large\mathfrak{~hope ~this ~help~you~~}}}}$

Answer 2

Answer:

HEY MATE...

P(X) = KY^2 + 2Y - 3K

A - 1K , B - 2 , C - (-3)K

SUM OF ZEROES = -B/A = -2/K

PRODUCT OF ZEROES = C/A = -3

ACCORDING TO THE QUESTION

SUM = 2 ( PRODUCT )

-2/K = 2 x -3

-2/K = -6

CROSS MULTIPLYING

SO K = 1/3 ANSWER

HOPE IT HELPS !!!

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