Question

square root of (7-4√3)​

Answer 1

$7-4\sqrt{3} \\ \\ \\ 4+3-4\sqrt{3} \\ \\ \\ 2^2+(\sqrt{3})^2-(2 \times 2 \times \sqrt{3})$

Now, 7 - 4√3 is in the form of a² + b² - 2ab, thus the square root is (a - b)² or (b - a)².

So,

$2^2+(\sqrt{3})^2-(2 \times 2 \times \sqrt{3}) \\ \\ \\ (2-\sqrt{3})^2\ \ \ \ \ \ \ \ \ \ \boxed{OR}\ \ \ \ \ \ \ \ \ \ (\sqrt{3}-2)^2$

Hence the square root is  2 - √3  or  √3 - 2.