Question

find the value of k such that the polynomial x^2-(k+6)x+2(2k-1) has sum of its zeros equal to one third of their product​

Answer 1

Answer:

given quadratic polynomial, x² - (k+6)x + 2(2k+1) = 0By comparing it with ax²+bx+c = 0,we geta = 1 , b = -(k+6) , c = 2(2k+1)Sum of zeroes = -b/a = -[-(k+6)]/1=… ... x2-(k+6)+2(2k+1) has sum of the zeros is half of their product. ... You can specify conditions of storing and accessing cookies in your browser.

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

Let α and β are the roots of given quadratic equation x² - ( k +6)x + 2(2k +1) = 0

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