Question

The value of k such that the quadratic polynomial x^2 - (k+6)x+2(2k+1) has sum of the zeroes as half of their product, is

Answer 1

Answer:

5

Step-by-step explanation:

x^2 - (k+6)x + 2(2k+1)

a = 1

b = -(k+b)

c = 2(2k+1)

alpha+b = -b/a = -{-(k+6)}/1 = k+6

alphaB = c/a = 2(2k+1)/1 = 2(2k+1)

alpha+B = alphaB/2

k+6 = 2(2k+1)/2

k+6 = 2k+1

5 = k