The set of values of p for which `x^2-2px+3p+4` is negative for atleast one real x is

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem of finding the set of values of \( p \) for which the quadratic expression \( x^2 - 2px + 3p + 4 \) is negative for at least one real \( x \), we will follow these steps:

### Step 1: Identify the quadratic expression
The given quadratic expression is:
\[ f(x) = x^2 - 2px + (3p + 4) \]

### Step 2: Determine the condition for negativity
For the quadratic expression to be negative for at least one real \( x \), the graph of the quadratic must intersect the x-axis at two points. This means that the discriminant \( D \) of the quadratic must be greater than zero.

### Step 3: Calculate the discriminant
The discriminant \( D \) for a quadratic equation \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = -2p \), and \( c = 3p + 4 \). Thus, we have:
\[ D = (-2p)^2 - 4(1)(3p + 4) \]
\[ D = 4p^2 - 4(3p + 4) \]
\[ D = 4p^2 - 12p - 16 \]

### Step 4: Set the discriminant greater than zero
To find the values of \( p \) for which the quadratic is negative for at least one real \( x \), we set the discriminant greater than zero:
\[ 4p^2 - 12p - 16 > 0 \]
Dividing the entire inequality by 4 gives:
\[ p^2 - 3p - 4 > 0 \]

### Step 5: Factor the quadratic inequality
Next, we factor the quadratic expression:
\[ p^2 - 3p - 4 = (p - 4)(p + 1) \]
Thus, we need to solve the inequality:
\[ (p - 4)(p + 1) > 0 \]

### Step 6: Determine the critical points
The critical points from the factors are \( p = 4 \) and \( p = -1 \). We will analyze the sign of the expression in the intervals determined by these points.

### Step 7: Test intervals
We will test the intervals:
1. \( (-\infty, -1) \)
2. \( (-1, 4) \)
3. \( (4, \infty) \)

- For \( p < -1 \) (e.g., \( p = -2 \)):
\((p - 4)(p + 1) = (-2 - 4)(-2 + 1) = (-6)(-1) > 0\)

- For \( -1 < p < 4 \) (e.g., \( p = 0 \)):
\((p - 4)(p + 1) = (0 - 4)(0 + 1) = (-4)(1) < 0\)

- For \( p > 4 \) (e.g., \( p = 5 \)):
\((p - 4)(p + 1) = (5 - 4)(5 + 1) = (1)(6) > 0\)

### Step 8: Write the solution
The expression \( (p - 4)(p + 1) > 0 \) is satisfied in the intervals:
\[ (-\infty, -1) \cup (4, \infty) \]

### Final Answer
Thus, the set of values of \( p \) for which \( x^2 - 2px + 3p + 4 \) is negative for at least one real \( x \) is:
\[ p \in (-\infty, -1) \cup (4, \infty) \]