Answer :
Answer:
[tex]k=\frac{-11}{2}[/tex].
Step-by-step explanation:
We are given [tex]\alpha[/tex] and [tex]\beta[/tex] are zeros of the polynomial [tex]x^2-(k+6)x+2(2k-1)[/tex].
We want to find the value of [tex]k[/tex] if [tex]\alpha+\beta=\frac{1}{2}[/tex].
Lets use veita's formula.
By that formula we have the following equations:
[tex]\alpha+\beta=\frac{-(-(k+6))}{1}[/tex] (-b/a where the quadratic is ax^2+bx+c)
[tex]\alpha \cdot \beta=\frac{2(2k-1)}{1}[/tex] (c/a)
Let's simplify those equations:
[tex]\alpha+\beta=k+6[/tex]
[tex]\alpha \cdot \beta=4k-2[/tex]
If [tex]\alpha+\beta=k+6[/tex] and [tex]\alpha+\beta=\frac{1}{2}[/tex], then [tex]k+6=\frac{1}{2}[/tex].
Let's solve this for k:
Subtract 6 on both sides:
[tex]k=\frac{1}{2}-6[/tex]
Find a common denominator:
[tex]k=\frac{1}{2}-\frac{12}{2}[/tex]
Simplify:
[tex]k=\frac{-11}{2}[/tex].