If $$x^2-4x+4b=0$$​has two real solutions, find the value of 'b'.

Correct option is D

Given:

x²- 4x + 4b = 0 has two real solutions.

Concept Used:

Discriminant for a quadratic equation ax² + bx + c = 0

b² - 4ac ≥ 0

​Solution:

The discriminant Δ for this equation is:

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

For the quadratic equation to have two real solutions, the discriminant should be non-negative:

​$$16 - 16b \geq 0 \\$$​​

Simplifying this inequality:

​$$\frac{16}{1} \geq \frac{16b}{1} \\1 \geq b \\$$​​

​$$\text{Thus, the value of }b\text{ must satisfy }b\le 1.$$

Therefore, the correct answer is b≤1 for the quadratic equation to have two real solutions.​