For this case we have the following polynomial:
\(x^2 - \frac{1}{4}x + k\)
We suppose that we have a standard equation of the form:
\(ax ^ 2 + bx + c\)
To make a perfect square trinomial, we need to complete the square.
Therefore, the value of c is:
\(c = (\frac{b}{2})^2\)
Using this definition for this case, we have:
\(k = (\frac{\frac{-1}{4}}{2})^2\)
\(k= ( \frac{-1}{8} )^2\)
\(k = \frac{1}{64}\)
Therefore, the perfect square trinomial is:
\(x^2 - \frac{1}{4}x + \frac{1}{64}\)
Answer:
The value of k is:
\(k = \frac{1}{64}\)
