Irrational Numbers

$\sqrt{7-4\sqrt{3}}$ Rewriting 7 as 4 + 3, we have: $\sqrt{4+3-4\sqrt{3}}$ By rewriting 3 in another form: $\sqrt{4+(\sqrt{3})^{2}-4\sqrt{3}}$ Thus, by rewriting the above expression as a square, we arrive at: $\sqrt{(2-\sqrt{3})^{2}}$ $\therefore 2-\sqrt{3}$

Using the first letter, then $\displaystyle \sqrt { 7 - 4 \sqrt { 3 } } = \sqrt { 7 -4 \times 1.7 } = \sqrt { 7 - 6.8 } = \sqrt { 0.2 }$ .

We can find the number easily which is $0.2$ .

$1 + 1.7 \ne 0.2$ , $1 - 1.7 \ne 0.2$ , $2 + 1.7 \ne 0.2$ , and $2 - 1.7 \equiv 0.2$ .

So, the answer is $2 - \sqrt { 3 }$ .