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: $m \in (-\infty, -8] \cup [0, \infty)$

Would you like more details or have any questions?

Here are