To determine the value(s) of $m$ for which the quadratic equation $0 = (3m + 4)x^2 + mx - 0.5$ has no real roots, we need to look at the discriminant of the quadratic equation. For a general quadratic equation of the form $ax^2 + bx + c = 0$, the discriminant $\Delta$ is given by: $\Delta = b^2 - 4ac$ If $\Delta < 0$, the quadratic equation has no real roots.

Step 1: Identify $a$, $b$, and $c$

From the equation: $0 = (3m + 4)x^2 + mx - 0.5$ we have:

• $a = 3m + 4$
• $b = m$
• $c = -0.5$

Step 2: Set up the Discriminant Condition

To ensure there are no real roots, we need $\Delta < 0$. Substituting $a$, $b$, and $c$ into the discriminant: $\Delta = b^2 - 4ac = m^2 - 4(3m + 4)(-0.5)$ Simplify this expression: $\Delta = m^2 + 2(3m + 4)$ $\Delta = m^2 + 6m + 8$

Step 3: Set the Inequality

We want: $m^2 + 6m + 8 < 0$ This is a quadratic inequality. We’ll first solve the corresponding equation: $m^2 + 6m + 8 = 0$ Using the quadratic formula: $m = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$ $m = \frac{-6 \pm \sqrt{36 - 32}}{2}$ $m = \frac{-6 \pm \sqrt{4}}{2}$ $m = \frac{-6 \pm 2}{2}$ So, $m = -2 \quad \text{and} \quad m = -4$

Step 4: Test the Interval for $m$

The inequality $m^2 + 6m + 8 < 0$ is satisfied for values of $m$ between the roots. Therefore: $-4 < m < -2$

Conclusion

The quadratic equation has no real roots for values of $m$ in the interval: $-4 < m < -2$

Would you like further details on this solution, or do you have any other questions?

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Here are some additional related questions you might explore:

1. How does the discriminant help determine the nature of the roots?
2. Can you find values of $m$ that would yield exactly one real root?
3. How would the answer change if the constant term were different?
4. What values of $m$ make the equation factorable?
5. How does the solution change if $m$ is constrained to integer values?

Tip: When solving inequalities involving a quadratic, factor the expression and analyze the intervals between roots to find where the inequality holds.