The least integral value of 'k' for which `(k -2)x^2 +8x+k+4>0` for all `xepsilonR`, is:

Options

A. 5
B. 4
C. 3
D. none of these

Explanation

To solve the inequality \((k - 2)x^2 + 8x + (k + 4) > 0\) for all \(x \in \mathbb{R}\), we need to ensure that the quadratic expression is always positive. This requires two conditions to be satisfied:

1. The coefficient of \(x^2\) must be positive.
2. The discriminant of the quadratic must be negative (to ensure there are no real roots).

### Step 1: Ensure the coefficient of \(x^2\) is positive
The coefficient of \(x^2\) is \(k - 2\). For the quadratic to open upwards, we need:
\[ k - 2 > 0 \]
This simplifies to:
\[ k > 2 \]

**Hint:** Check the sign of the coefficient of \(x^2\) to determine if the parabola opens upwards.

### Step 2: Ensure the discriminant is negative
The discriminant \(D\) of a quadratic \(ax^2 + bx + c\) is given by:
\[ D = b^2 - 4ac \]
For our quadratic, \(a = k - 2\), \(b = 8\), and \(c = k + 4\). Thus, the discriminant is:
\[ D = 8^2 - 4(k - 2)(k + 4) \]
Calculating this gives:
\[ D = 64 - 4[(k - 2)(k + 4)] \]
Expanding the product:
\[ (k - 2)(k + 4) = k^2 + 4k - 2k - 8 = k^2 + 2k - 8 \]
So, substituting back into the discriminant:
\[ D = 64 - 4(k^2 + 2k - 8) \]
This simplifies to:
\[ D = 64 - 4k^2 - 8k + 32 \]
\[ D = 96 - 4k^2 - 8k \]
We want this to be less than 0:
\[ 96 - 4k^2 - 8k < 0 \]
Dividing the entire inequality by -4 (and reversing the inequality sign):
\[ k^2 + 2k - 24 > 0 \]

**Hint:** Use the discriminant to determine the conditions for the quadratic to be positive.

### Step 3: Factor the quadratic
Now, we factor \(k^2 + 2k - 24\):
\[ k^2 + 2k - 24 = (k + 6)(k - 4) \]
We need to find the intervals where this product is greater than zero. The critical points are \(k = -6\) and \(k = 4\).

**Hint:** Identify the intervals based on the roots of the quadratic.

### Step 4: Test intervals
The intervals to test are:
- \( (-\infty, -6) \)
- \( (-6, 4) \)
- \( (4, \infty) \)

Testing each interval:
1. For \(k < -6\), say \(k = -7\):
\[ (-7 + 6)(-7 - 4) = (-1)(-11) > 0 \]
2. For \(-6 < k < 4\), say \(k = 0\):
\[ (0 + 6)(0 - 4) = (6)(-4) < 0 \]
3. For \(k > 4\), say \(k = 5\):
\[ (5 + 6)(5 - 4) = (11)(1) > 0 \]

Thus, the solution set for \(k\) is:
\[ k \in (-\infty, -6) \cup (4, \infty) \]

### Step 5: Combine conditions
From Step 1, we have \(k > 2\). The intersection of \(k > 2\) and \(k \in (4, \infty)\) gives:
\[ k \in (4, \infty) \]

### Step 6: Find the least integral value of \(k\)
The least integral value of \(k\) that satisfies \(k > 4\) is:
\[ k = 5 \]

### Final Answer:
The least integral value of \(k\) for which \((k - 2)x^2 + 8x + (k + 4) > 0\) for all \(x \in \mathbb{R}\) is:
\[ \boxed{5} \]