For what values of a is the inequality ` (x^(2) +ax-2)/( x^(2) -x+1) lt 2` satisfied for all real values of x?

Answer

To solve the inequality \(\frac{x^2 + ax - 2}{x^2 - x + 1} < 2\) for all real values of \(x\), we will follow these steps:

### Step 1: Analyze the Denominator
First, we need to ensure that the denominator \(x^2 - x + 1\) is always positive for all real \(x\).

The discriminant \(D\) of the quadratic \(x^2 - x + 1\) is given by:
\[ D = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic \(x^2 - x + 1\) does not have real roots and opens upwards (as the coefficient of \(x^2\) is positive). Therefore, \(x^2 - x + 1 > 0\) for all real \(x\).

**Hint:** Check the discriminant to determine if a quadratic is always positive.

### Step 2: Cross-Multiply the Inequality
Since the denominator is always positive, we can safely multiply both sides of the inequality by \(x^2 - x + 1\) without changing the inequality:
\[ x^2 + ax - 2 < 2(x^2 - x + 1) \]

### Step 3: Simplify the Inequality
Now, simplify the right-hand side:
\[ x^2 + ax - 2 < 2x^2 - 2x + 2 \]
Rearranging gives:
\[ x^2 + ax - 2 - 2x^2 + 2x - 2 < 0 \]
This simplifies to:
\[ -x^2 + (a + 2)x - 4 < 0 \]
or equivalently:
\[ x^2 - (a + 2)x + 4 > 0 \]

### Step 4: Analyze the New Quadratic
For the quadratic \(x^2 - (a + 2)x + 4\) to be positive for all \(x\), its discriminant must be negative:
\[ D = (-(a + 2))^2 - 4(1)(4) < 0 \]
This simplifies to:
\[ (a + 2)^2 - 16 < 0 \]
or:
\[ (a + 2)^2 < 16 \]

### Step 5: Solve the Inequality
Taking the square root of both sides gives:
\[ -4 < a + 2 < 4 \]
Subtracting 2 from all parts results in:
\[ -6 < a < 2 \]

### Final Answer
Thus, the values of \(a\) for which the inequality \(\frac{x^2 + ax - 2}{x^2 - x + 1} < 2\) is satisfied for all real \(x\) are:
\[ a \in (-6, 2) \]