To determine the range of values of $m$ for which the equation $mx^2 + mx - 2 = 0$ has one or two real roots, we analyze the discriminant ($\Delta$) of the quadratic equation.

General Quadratic Equation and Discriminant:

For a quadratic equation $ax^2 + bx + c = 0$, the discriminant is: $\Delta = b^2 - 4ac$ The nature of the roots depends on $\Delta$:

1. $\Delta > 0$: Two distinct real roots.
2. $\Delta = 0$: One real root (repeated root).
3. $\Delta < 0$: No real roots (complex roots).

Step-by-Step Solution:

For the given equation $mx^2 + mx - 2 = 0$:

• Coefficient of $x^2$: $a = m$
• Coefficient of $x$: $b = m$
• Constant term: $c = -2$

The discriminant is: $\Delta = b^2 - 4ac = m^2 - 4(m)(-2)$

Simplify: $\Delta = m^2 + 8m$

Condition for One or Two Real Roots:

1. At least one real root exists if $\Delta \geq 0$: $m^2 + 8m \geq 0$

   Factorize: $m(m + 8) \geq 0$

   Solve the inequality:

   • The critical points are $m = 0$ and $m = -8$.
   • Test intervals around $m = 0$ and $m = -8$:
     • For $m < -8$: $m(m + 8) > 0$.
     • For $-8 < m < 0$: $m(m + 8) < 0$.
     • For $m > 0$: $m(m + 8) > 0$.

   The solution to $m(m + 8) \geq 0$ is: $m \in (-\infty, -8] \cup [0, \infty)$

Final Answer:

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

Would you like further details on solving quadratic inequalities or interpreting discriminants? Let me know!

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Related Questions:

1. How do you solve quadratic equations when $\Delta = 0$?
2. What is the significance of the sign of $a$ in quadratic equations?
3. How do you graphically represent quadratic inequalities like $m(m + 8) \geq 0$?
4. What happens to the roots if $\Delta < 0$?
5. How can the vertex form of a quadratic equation help in analyzing roots?

Tip:

Always check the coefficient of $x^2$ ($a$) in a quadratic equation—if $a = 0$, the equation becomes linear!