Find the rational number k satisfying −√12/2 − √27/9 + √75/15 + √108 = k√3.

Step-by-step solution

1. 1

   Reduce every radical to a multiple of √3. Each radicand is $3$ times a perfect square, so all four terms are commensurable — which is exactly why the sum can be written as $k\sqrt3$:

   $\sqrt{12}=2\sqrt3,\quad \sqrt{27}=3\sqrt3,\quad \sqrt{75}=5\sqrt3,\quad \sqrt{108}=6\sqrt3$

2. 2

   Substitute into the expression.

   $-\frac{2\sqrt3}{2}-\frac{3\sqrt3}{9}+\frac{5\sqrt3}{15}+6\sqrt3$

3. 3

   Simplify each coefficient.

   $-\sqrt3-\frac{\sqrt3}{3}+\frac{\sqrt3}{3}+6\sqrt3$

   The two middle terms are exact negatives of each other and cancel — a designed feature of the problem, and a useful reminder to simplify before finding a common denominator.

4. 4

   Collect the surviving terms.

   $(-1+6)\sqrt3=5\sqrt3$

5. 5

   Read off k.

   $\boxed{k=5}$

6. 6

   Verify numerically. With $\sqrt3\approx 1.7320508$: $-\tfrac{3.4641}{2}=-1.7321$, $-\tfrac{5.1962}{9}=-0.5774$, $+\tfrac{8.6603}{15}=+0.5774$, $+10.3923$. The sum is $8.6603=5\sqrt3$ ✓.