Question

Find the value of k such that the quadratic
polynomial x²- (k + 6)x + 2 (2k + 1) has sum
of the zeroes as double of their product.​

Answer 1

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