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 = ${\displaystyle \frac{1}{2}}$ product of zeroes

⇒ ${\displaystyle \frac{-[-(\text{k}+6)]}{1}=\frac{1}{2}\times \frac{2(2\text{k}-1)}{1}}$

⇒ ${\displaystyle \text{k}+6=2\text{k}-1}$
⇒ ${\displaystyle 6+1=2\text{k}-\text{k}}$
⇒ ${\displaystyle \text{k}=7}$