Question

Find the value of k such that the polynomial x2 − (k + 6)x + 2(2k −1) has sum of its zeros equal to half of their product.

Answer 1

Given polynomial is x² - (k + 6) x + 2 (2k – 1)

Here
a = 1, b = - (k + 6), c =  2 (2k – 1)

Given that,
Sum of zeroes = `(1)/(2)` product of zeroes

⇒ `[-[-("k"+6)]]/(1) = (1)/(2) xx (2(2"k" - 1))/(1)`

⇒ `"k" + 6 = 2"k" - 1`
⇒ `6 + 1 = 2"k" - "k"`
⇒ `"k" = 7`