To solve for the real number $m \in \mathbb{R}$ such that the given conditions hold, let's break the problem into two parts and solve step-by-step:

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1. Condition 1: The equation $2x^2 - 5x + 2m = 0$ has complex solutions.

For a quadratic equation $ax^2 + bx + c = 0$ to have complex solutions, its discriminant ($\Delta$) must satisfy:

$\Delta = b^2 - 4ac < 0$

Here: $a = 2, \quad b = -5, \quad c = 2m$

The discriminant is: $\Delta = (-5)^2 - 4(2)(2m) = 25 - 16m$

For complex solutions: $25 - 16m < 0 \quad \Rightarrow \quad m > \frac{25}{16}$

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2. Condition 2: The equation $(m+5)x^2 + 3x - 1 = 0$ has a double solution.

For a quadratic equation to have a double solution, its discriminant ($\Delta$) must equal zero:

$\Delta = b^2 - 4ac = 0$

Here: $a = m+5, \quad b = 3, \quad c = -1$

The discriminant is: $\Delta = 3^2 - 4(m+5)(-1) = 9 + 4(m+5)$

For a double solution: $9 + 4(m+5) = 0 \quad \Rightarrow \quad 4(m+5) = -9 \quad \Rightarrow \quad m+5 = -\frac{9}{4}$

Solving for $m$: $m = -\frac{9}{4} - 5 = -\frac{9}{4} - \frac{20}{4} = -\frac{29}{4}$

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3. Combine the conditions:

From Condition 1: $m > \frac{25}{16}$.

From Condition 2: $m = -\frac{29}{4}$.

There is no $m$ that satisfies both conditions simultaneously.

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Conclusion:

There is no real value of $m$ such that both conditions hold true. The problem has no solution in $\mathbb{R}$.

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Follow-up Questions:

1. What happens if we modify the first condition to allow real roots?
2. How does the discriminant determine the nature of roots for quadratic equations?
3. What if the second equation has distinct real roots instead of a double solution?
4. Can we visualize the solution space for $m$ in both conditions graphically?
5. How can we derive similar conditions for cubic equations?

Tip:

When dealing with quadratic equations, always start by analyzing the discriminant to determine the nature of the roots.