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The question asks to "find the range of values of \( m \) for which the quadratic equation \( mx^2 + mx - 2 = 0 \) has one or two real roots."

### Solution:

The general quadratic equation is:
\[
ax^2 + bx + c = 0
\]
For the equation \( mx^2 + mx - 2 = 0 \), the coefficients are:
- \( a = m \)
- \( b = m \)
- \( c = -2 \)

To find the nature of the roots (real, equal, or imaginary), we use the **discriminant** \( \Delta \) formula:
\[
\Delta = b^2 - 4ac
\]
In this case:
\[
\Delta = m^2 - 4(m)(-2)
\]
\[
\Delta = m^2 + 8m
\]

For the quadratic equation to have **one or two real roots**, the discriminant must be non-negative:
\[
\Delta \geq 0
\]
\[
m^2 + 8m \geq 0
\]

### Solving the inequality:

Factor the expression:
\[
m(m + 8) \geq 0
\]

This quadratic inequality can be solved by analyzing the critical points, \( m = 0 \) and \( m = -8 \), which divide the number line into intervals. We test each interval to determine where the product is non-negative.

1. **For \( m < -8 \)**: Both factors are negative, so the product is positive.
2. **For \( -8 < m < 0 \)**: One factor is negative and the other is positive, so the product is negative.
3. **For \( m > 0 \)**: Both factors are positive, so the product is positive.

Thus, the quadratic inequality holds when \( m \in (-\infty, -8] \cup [0, \infty) \).

### Conclusion:
The quadratic equation \( mx^2 + mx - 2 = 0 \) has one or two real roots for:
\[
m \in (-\infty, -8] \cup [0, \infty)
\]

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Find the range of values of m for which the quadratic equation mx^2 + mx - 2 = 0 has one or two real roots.

Find the range of values for m where the quadratic equation mx^2 + mx - 2 = 0 has one or two real roots. Learn how to apply the discriminant formula and solve inequalities. Suitable for students in Grades 9-10.

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