Unraveling 4x² + 5x + C = 0: Finding the Unsolvable Condition

What does $\square$ need to be so that the equation below has no solution?

$4x^2+5x+\square=0$

To solve this problem, we need to find the value of $C$ in the quadratic equation $4x^2 + 5x + C = 0$ such that the equation has no real solution. This occurs when the discriminant of the quadratic equation is less than zero.

The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by:

$\Delta = b^2 - 4ac$

In our equation:

• $a = 4$
• $b = 5$
• $c = C$

The discriminant becomes:

$\Delta = 5^2 - 4 \cdot 4 \cdot C = 25 - 16C$

For the quadratic equation to have no real solutions, the discriminant must be less than zero:

$25 - 16C < 0$

Solving this inequality for $C$:

$25 < 16C$

$\frac{25}{16} < C$

Therefore, the condition for $C$ is that it must be greater than $\frac{25}{16}$ for the quadratic equation to have no real solutions.

Therefore, the correct answer is $\frac{25}{16} < C$.