Question

If the sum of zeros of quadratic polynomial kx2 - 2 x -3k is equal to the twice their product find the value of k

Answer 1

Heya !!






KX² - 2X - 3K




Here,



A = K , B = -2 and C = -3K.





Sum of zeroes = -B/A


Alpha + Beta = 2/K





And,


Product of zeroes = C/A


Alpha × Beta = -3K/K = -3





According to the question,



Sum of zeroes = 2 × Product of zeroes

2/k = 2 × -3




-6K = 2





K = -1/3

Answer 2

Concept

For a quadratic equation

a$x^{2}$ - 2b - c

sum of roots = -b/a

Product of roots = c/a

Given

a polynomial k$x^{2}$ - 2 x -3k

whose sum of zeroes is equal to the twice their product

Find

We have to find the value of k

Solution

We have,

k$x^{2}$ - 2x - 3k = 0

We know that sum of zeroes = -b/a

Therefore,

sum = -(-2)/k

sum = 2/k

Product of zeroes = c/a

Product = -3k/k

Product = -3

Also,

Sum of zeroes = 2(product of zeroes)

⇒ 2/k = 2 x (-3)

2/k = -6

k = -1/3

Thus the value of k is -1/3

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