To determine the value of $b$ for which the quadratic equation $x^2 - 4x + 4b = 0$ has two real solutions, we can use the discriminant method. The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by:

$\Delta = b^2 - 4ac$

For the equation to have real solutions, the discriminant must be non-negative, i.e., $\Delta \geq 0$.

Given the equation:

$x^2 - 4x + 4b = 0$

Here, $a = 1$, $b = -4$, and $c = 4b$.

Now, calculate the discriminant:

$\Delta = (-4)^2 - 4(1)(4b) = 16 - 16b$

For the equation to have real solutions, we need:

$\Delta \geq 0 \implies 16 - 16b \geq 0$

$16 \geq 16b \implies b \leq 1$

Thus, for the quadratic equation to have two real solutions, $b$ must satisfy $b \leq 1$.

Therefore, the correct answer is:

Option D) $b \geq 1$ does not hold.
The correct range is $b \leq 1$.
But since it was phrased in reverse, correct understanding would imply the only answer covering $b = 1$ and $b < 1$ is B).

Do you want further details or have any questions?

Here are 5 related questions:

1. What are the different conditions on the discriminant for a quadratic equation to have distinct real, repeated, or complex solutions?
2. How does the vertex form of a quadratic equation relate to its discriminant?
3. Can a quadratic equation have exactly one real solution? Under what circumstances?
4. What happens to the roots of the quadratic equation when $b = 1$ in this problem?
5. How would the graph of $x^2 - 4x + 4b = 0$ change as $b$ varies?

Tip: Always remember that the discriminant determines the nature of the roots of a quadratic equation.