B3. Solve the irrational equation $\sqrt{x^{2}+4}-\sqrt{4 x}=0$.
B3. First, we rearrange the equation to $\mathrm{v} \sqrt{x^{2}+4}=\sqrt{4 x}$ and square both sides: $x^{2}+4=4 x$. When we move the term $4 x$ to the left side of the equation, we get $x^{2}-4 x+4=0$, which we can write in the form: $(x-2)^{2}=0$. The only solution is $x=2$. We also perform a check: $\sqrt{2^{2}+4}-\sqrt{4 \cdot 2}=0$, which confirms that $x=2$ solves the given irrational equation.

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-13.jpg?height=57&width=1641&top_left_y=454&top_left_x=276)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-13.jpg?height=60&width=1641&top_left_y=501&top_left_x=276)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-13.jpg?height=55&width=1637&top_left_y=555&top_left_x=278)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-13.jpg?height=60&width=1641&top_left_y=604&top_left_x=276)

![](https://cdn.mathpix.com/cropped/2024_06_07_8d4f73ada7862f989e67g-13.jpg?height=49&width=1642&top_left_y=661&top_left_x=273)

Completed check..............................................................................................