(i) If `x^2-px+4` >0 for all real 'x' then find 'p' ?

Answer

To solve the problem, we need to determine the value of \( p \) such that the quadratic expression \( x^2 - px + 4 \) is greater than zero for all real values of \( x \).

### Step-by-Step Solution:

1. **Identify the Quadratic Form**:
The given quadratic expression is \( x^2 - px + 4 \). This can be compared to the standard form of a quadratic equation \( ax^2 + bx + c \), where:
- \( a = 1 \)
- \( b = -p \)
- \( c = 4 \)

2. **Condition for the Quadratic to be Positive**:
For the quadratic expression \( ax^2 + bx + c \) to be greater than zero for all real \( x \), the following conditions must be satisfied:
- The coefficient \( a \) must be greater than zero (which it is, since \( a = 1 > 0 \)).
- The discriminant \( D \) must be less than zero. The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]

3. **Calculate the Discriminant**:
Substituting the values of \( a \), \( b \), and \( c \) into the discriminant formula:
\[ D = (-p)^2 - 4 \cdot 1 \cdot 4 \]
Simplifying this gives:
\[ D = p^2 - 16 \]

4. **Set the Discriminant Less Than Zero**:
To ensure the quadratic is always positive, we set the discriminant less than zero:
\[ p^2 - 16 < 0 \]

5. **Solve the Inequality**:
Rearranging the inequality:
\[ p^2 < 16 \]
Taking the square root of both sides gives:
\[ -4 < p < 4 \]

6. **Conclusion**:
The values of \( p \) that satisfy the condition are:
\[ p \in (-4, 4) \]

### Final Answer:
The value of \( p \) must lie in the interval \( (-4, 4) \).