The values of " $p$ " in the equation $p x^{2}-7 x+8=0$ that will produce no real roots are $p= \pm \frac{49}{32}$ $p>-\frac{49}{32}$ $p>\frac{49}{32}$ $p<\frac{49}{32}$

Step 1 ：The roots of a quadratic equation \(ax^2 + bx + c = 0\) are given by the formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). The term under the square root, \(b^2 - 4ac\), is called the discriminant. If the discriminant is less than zero, the roots are not real. So, we need to find the values of \(p\) for which the discriminant of the given equation is less than zero.

Step 2 ：Let's calculate the discriminant of the given equation. The discriminant is \(49 - 32p\).

Step 3 ：We need to find the values of \(p\) for which the discriminant is less than zero. This gives us the inequality \(49 - 32p < 0\).

Step 4 ：Solving this inequality, we find that the solution is \(p > \frac{49}{32}\). This means that for values of \(p\) greater than \(\frac{49}{32}\), the equation will have no real roots.

Step 5 ：Final Answer: \(\boxed{p>\frac{49}{32}}\)