Solving Quadratic Inequalities

The question asks for the condition on the value of 'p' such that the quadratic expression \(x^2 - px + 4\) is greater than zero for all real values of 'x'.

A quadratic expression of the form \(ax^2 + bx + c\) is positive for all real values of 'x' if and only if two conditions are met:

• The coefficient of \(x^2\), which is 'a', must be positive (i.e., \(a > 0\)).
• The discriminant of the quadratic equation \(ax^2 + bx + c = 0\), denoted by \(\Delta\), must be negative (i.e., \(\Delta < 0\)). The discriminant is calculated as \(\Delta = b^2 - 4ac\).

Analyzing the Given Quadratic Expression

The given quadratic expression is \(x^2 - px + 4\). Comparing this with the standard form \(ax^2 + bx + c\), we can identify the coefficients:

• \(a = 1\)
• \(b = -p\)
• \(c = 4\)

Applying the Conditions

First condition: \(a > 0\)

In our case, \(a = 1\), and \(1 > 0\). This condition is satisfied.

Second condition: \(\Delta < 0\)

The discriminant is \(\Delta = b^2 - 4ac\). Substituting the values of a, b, and c:

\[ \Delta = (-p)^2 - 4(1)(4) \]

\[ \Delta = p^2 - 16 \]

For the quadratic expression to be always positive, the discriminant must be less than zero:

\[ p^2 - 16 < 0 \]

Solving the Inequality for p

We need to solve the inequality \(p^2 - 16 < 0\).

We can factor the left side as a difference of squares:

\[ (p - 4)(p + 4) < 0 \]

For the product of two factors to be negative, one factor must be positive and the other must be negative. There are two possible cases:

Case 1: \(p - 4 > 0\) and \(p + 4 < 0\)

• \(p > 4\)
• \(p < -4\)

This case requires 'p' to be both greater than 4 and less than -4 simultaneously, which is impossible.

Case 2: \(p - 4 < 0\) and \(p + 4 > 0\)

• \(p < 4\)
• \(p > -4\)

This case requires 'p' to be greater than -4 and less than 4. This can be written as \(-4 < p < 4\).

Alternatively, from \(p^2 < 16\), taking the square root of both sides gives \(|p| < \sqrt{16}\), which simplifies to \(|p| < 4\). The inequality \(|p| < 4\) is equivalent to \(-4 < p < 4\).

So, the condition for \(x^2 - px + 4 > 0\) for all real values of x is \(|p| < 4\).

Comparing with Options

Let's look at the given options:

Our derived condition is \(|p| < 4\), which matches Option 1.

Revision Table: Quadratic Expression Conditions

Additional Information: Understanding the Discriminant

The discriminant, \(\Delta = b^2 - 4ac\), tells us about the nature of the roots of a quadratic equation \(ax^2 + bx + c = 0\), and consequently, the behavior of the quadratic function \(f(x) = ax^2 + bx + c\).

• If \(\Delta > 0\): The quadratic equation has two distinct real roots. The graph of the function intersects the x-axis at two different points. The function changes sign at these roots.
• If \(\Delta = 0\): The quadratic equation has exactly one real root (or two equal real roots). The graph of the function touches the x-axis at exactly one point. The function does not change sign but is zero at the root.
• If \(\Delta < 0\): The quadratic equation has no real roots; it has two complex conjugate roots. The graph of the function does not intersect the x-axis. This means the function is either always positive (if \(a > 0\)) or always negative (if \(a < 0\)).

For the expression \(x^2 - px + 4\) to be always positive, its graph must be a parabola opening upwards (which is true since \(a=1 > 0\)) and must never touch or cross the x-axis. This exactly corresponds to the case where the discriminant is negative (\(\Delta < 0\)).