Find the value of p for which \(x^{2} - x + p\) becomes a perfect square.
Answer Details
To make the expression \(x^2-x+p\) a perfect square, we need to add and subtract a suitable constant term.
To do this, we can start by finding the square of half the coefficient of the linear term. That is, we find the square of \(\frac{1}{2}\) which is \(\frac{1}{4}\).
So, we have:
$$x^2-x+\frac{1}{4}-\frac{1}{4}+p$$
Now, we can write this expression as a perfect square by factoring the first three terms:
$$\left(x-\frac{1}{2}\right)^2+p-\frac{1}{4}$$
For this expression to be a perfect square, the last term must be zero, so we have:
$$p-\frac{1}{4}=0$$
Solving for \(p\), we get:
$$p=\frac{1}{4}$$
Therefore, the value of \(p\) that makes \(x^2-x+p\) a perfect square is \(\frac{1}{4}\).