Simplify sqrt(3 + sqrt 5) + sqrt(3 - sqrt 5).

Step-by-step solution

1. 1

   Check the expression is real, then name it. Since $\sqrt5\approx 2.236<3$, both $3+\sqrt5$ and $3-\sqrt5$ are positive, so both square roots are real and positive. Set

   $S=\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}$

   Naming the sum is what makes the next move possible: nested radicals rarely simplify term by term, but their square often does.

2. 2

   Square the sum with $(a+b)^2=a^2+b^2+2ab$. Each square root disappears when squared:

   $S^2=(3+\sqrt5)+(3-\sqrt5)+2\sqrt{(3+\sqrt5)(3-\sqrt5)}$

3. 3

   Simplify the two easy pieces. The $\sqrt5$ terms cancel in the sum, and the product under the remaining radical is a difference of squares:

   $(3+\sqrt5)+(3-\sqrt5)=6,\qquad (3+\sqrt5)(3-\sqrt5)=3^2-(\sqrt5)^2=9-5=4$

   This cancellation is the whole reason the conjugate pair was chosen.

4. 4

   Assemble $S^2$.

   $S^2=6+2\sqrt4=6+2\cdot 2=10$

5. 5

   Choose the correct root. Squaring can introduce a spurious sign, so decide it from the original expression: $S$ is a sum of two positive numbers, hence $S>0$ and

   $S=\sqrt{10}$

   Numerically $\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}=2.2882+0.8740=3.1623=\sqrt{10}$. ✓