Question

The question is in the attached file

Answer 1

Answer:

B.  √5

Step-by-step explanation:

Given:

$K=\sqrt{9-4\sqrt{5}}$

$\implies K+2=\sqrt{9-4\sqrt{5}}+2$

Rewrite 9 as 5 + 4:

$= \sqrt{5-4\sqrt{5}+4}+2$

Rewrite 5 as (√5)², 4 as 2² and 4√5 as 2 · 2√5:

$= \sqrt{(\sqrt{5})^2-2 \cdot 2\sqrt{5}+2^2}+2$

Apply the perfect square formula  $a^2-2ab+b^2=(a-b)^2$:

$=\sqrt{(\sqrt{5}-2)^2}+2$

Apply radical rule  $\sqrt{a^2}=a$

$=\sqrt{5}-2+2$

Simplify:

$=\sqrt{5}$