Question

find the value of k such that the quadratic polynomial x2-(k+6)+2(2k+1) has sum of the zeros is half of their product.

Answer 1

Let α and β are the roots of given quadratic equation x² - ( k +6)x + 2(2k +1) = 0 [ you did mistake in typing of equation , I just correct it ]

Now, sum of roots = α + β = - {-( k + 6)}/1 = (k + 6)
product of roots = αβ = 2(2k + 1)/1= 2(2k + 1)

A/C to question,
sum of roots ( zeros ) = 1/2 × products of roots zeros
⇒ (k + 6) = 1/2 × 2(2k + 1)
⇒ (k + 6) = (2k + 1)
⇒ k + 6 = 2k + 1
⇒ k = 5

Hence, k = 5

Answer 2

Given quadratic polynomial,
x² - (k+6)x + 2(2k+1) = 0

By comparing it with ax²+bx+c = 0,we get
a = 1 , b = -(k+6) , c = 2(2k+1)

Sum of zeroes = -b/a
= -[-(k+6)]/1
= k+6

Product of zeroes = c/a
= 2(2k+1)/1
= 2(2k+1)

As per problem,
Sum of zeroes = ½ × product of zeroes
k+6 = ½ × 2(2k+1)
k+6 = 2k+1
2k-k = 6-1
k = 5

Hope it helps