Expand and simplify the square root of 2 times (the square root of 7 plus 6 times the square root of 72).

Step-by-step solution

1. 1

   Distribute the outside radical over both terms. $\sqrt2\left(\sqrt7+6\sqrt{72}\right) = \sqrt2\cdot\sqrt7 + \sqrt2\cdot6\sqrt{72}.$ The coefficient $6$ is just carried along; only the radicals combine.

2. 2

   Multiply the first pair of radicals. Using $\sqrt a\,\sqrt b = \sqrt{ab}$ for nonnegative $a,b$: $\sqrt2\cdot\sqrt7 = \sqrt{14}.$ Since $14 = 2\cdot7$ has no square factor, $\sqrt{14}$ is already in simplest surd form.

3. 3

   Multiply the second pair - and notice it becomes rational. $\sqrt2\cdot\sqrt{72} = \sqrt{144} = 12,$ because $2\cdot72 = 144$ is a perfect square. This is the point of the question: one term stays irrational and one does not.

4. 4

   Apply the coefficient. $6 \cdot 12 = 72$, so the second term is the integer $72$.

5. 5

   Assemble the answer. $\sqrt2\left(\sqrt7+6\sqrt{72}\right) = \sqrt{14}+72 = 72+\sqrt{14}.$ The two terms are unlike (one rational, one a surd), so they cannot be combined any further.

6. 6

   Check numerically. $\sqrt2 \approx 1.414214$, $\sqrt7 \approx 2.645751$, $\sqrt{72} \approx 8.485281$, so the original is $1.414214(2.645751+50.911688) \approx 75.741657$; and $72+\sqrt{14} \approx 72+3.741657 = 75.741657$. They agree.

7. 7

   Note the alternative route. Simplifying $\sqrt{72} = 6\sqrt2$ first turns the second term into $\sqrt2\cdot36\sqrt2 = 36\cdot2 = 72$ - the same result with even less arithmetic.