To solve for [tex]\( k \)[/tex] given that [tex]\(\alpha\)[/tex] and [tex]\(1 / \alpha\)[/tex] are the zeroes of the quadratic polynomial [tex]\(2x^2 - x + 8k\)[/tex], we will use Vieta's formulas, which relate the coefficients of the polynomial to the sums and products of its roots.
The polynomial is given by:
[tex]\[ 2x^2 - x + 8k \][/tex]
1. Sum of the roots:
By Vieta's formulas, the sum of the roots of a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex] is given by:
[tex]\[
\alpha + \frac{1}{\alpha} = -\frac{b}{a}
\][/tex]
For our polynomial [tex]\(2x^2 - x + 8k\)[/tex]:
[tex]\[
a = 2, \quad b = -1, \quad c = 8k
\][/tex]
So the sum of the roots is:
[tex]\[
\alpha + \frac{1}{\alpha} = -\frac{-1}{2} = \frac{1}{2}
\][/tex]
2. Product of the roots:
By Vieta's formulas, the product of the roots is given by:
[tex]\[
\alpha \cdot \frac{1}{\alpha} = \frac{c}{a}
\][/tex]
For our polynomial:
[tex]\[
\alpha \cdot \frac{1}{\alpha} = \frac{8k}{2} = 4k
\][/tex]
But [tex]\(\alpha \cdot \frac{1}{\alpha} = 1\)[/tex]. Therefore:
[tex]\[
1 = 4k
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
k = \frac{1}{4}
\][/tex]
So, the value of [tex]\( k \)[/tex] is:
[tex]\[ \boxed{\frac{1}{4}} \][/tex]
