Consider the equation `(x^2 + x + 1)^2-(m-3)(x^2 + x + 1) +m=0--(1),` where `m` is a real parameter. By putting `x^2+x+1=t--(2)` then `t >= 3/4` for real `x` the equation can be transferred to `f(t) = t^2- (m-3)t + m=0` . At what values of m for which the equation (1) will have real roots?

Answer

To solve the problem, we need to determine the values of \( m \) for which the equation

\[ (x^2 + x + 1)^2 - (m-3)(x^2 + x + 1) + m = 0 \tag{1} \]

has real roots. We start by substituting \( t = x^2 + x + 1 \), which transforms the equation into

\[ f(t) = t^2 - (m-3)t + m = 0 \tag{2} \]

Given that \( t \geq \frac{3}{4} \) for real \( x \), we need to ensure that the quadratic equation \( f(t) = 0 \) has real roots.

### Step 1: Determine the condition for real roots

For the quadratic equation \( f(t) = t^2 - (m-3)t + m = 0 \) to have real roots, the discriminant must be non-negative:

\[ D = b^2 - 4ac \geq 0 \]

Here, \( a = 1 \), \( b = -(m-3) \), and \( c = m \). Thus, the discriminant \( D \) is:

\[ D = (-(m-3))^2 - 4 \cdot 1 \cdot m \]

### Step 2: Simplify the discriminant

Calculating \( D \):

\[ D = (m-3)^2 - 4m \]

Expanding this gives:

\[ D = m^2 - 6m + 9 - 4m = m^2 - 10m + 9 \]

### Step 3: Set up the inequality

We need to solve the inequality:

\[ m^2 - 10m + 9 \geq 0 \]

### Step 4: Factor the quadratic

Factoring the quadratic:

\[ m^2 - 10m + 9 = (m - 1)(m - 9) \]

### Step 5: Analyze the intervals

To find the intervals where the product is non-negative, we analyze the sign of \( (m - 1)(m - 9) \):

- The roots are \( m = 1 \) and \( m = 9 \).
- The critical points divide the number line into intervals: \( (-\infty, 1) \), \( (1, 9) \), and \( (9, \infty) \).

### Step 6: Test the intervals

1. **For \( m < 1 \)**: Choose \( m = 0 \):
\[ (0 - 1)(0 - 9) = 1 \cdot 9 > 0 \]
(Positive)

2. **For \( 1 < m < 9 \)**: Choose \( m = 5 \):
\[ (5 - 1)(5 - 9) = 4 \cdot (-4) < 0 \]
(Negative)

3. **For \( m > 9 \)**: Choose \( m = 10 \):
\[ (10 - 1)(10 - 9) = 9 \cdot 1 > 0 \]
(Positive)

### Step 7: Conclusion

The intervals where \( (m - 1)(m - 9) \geq 0 \) are:

\[ m \in (-\infty, 1] \cup [9, \infty) \]

Thus, the values of \( m \) for which the equation (1) has real roots are:

\[ \boxed{(-\infty, 1] \cup [9, \infty)} \]