Rationalize the denominator of 17 divided by (2 + the square root of 11).

Step-by-step solution

1. 1

   Pick the conjugate, not just any multiplier. The denominator $2+\sqrt{11}$ is a binomial with one irrational term. Multiplying it by its conjugate $2-\sqrt{11}$ is the only choice that kills the radical in one move, because $(a+b)(a-b)=a^2-b^2$ squares the $\sqrt{11}$ away. Multiplying top and bottom by the same quantity leaves the value unchanged:

   $\frac{17}{2+\sqrt{11}}\cdot\frac{2-\sqrt{11}}{2-\sqrt{11}}$

2. 2

   Clear the denominator with the difference of squares.

   $(2+\sqrt{11})(2-\sqrt{11})=2^2-(\sqrt{11})^2=4-11=-7$

   Note the sign: because $11>4$, the result is negative. That negative denominator is the step students most often mishandle.

3. 3

   Expand the numerator. Distribute the $17$:

   $17(2-\sqrt{11})=34-17\sqrt{11}$

   so the fraction is now

   $\frac{34-17\sqrt{11}}{-7}$

4. 4

   Move the minus sign to the numerator. A negative denominator is not considered simplified. Multiply numerator and denominator by $-1$:

   $\frac{34-17\sqrt{11}}{-7}=\frac{17\sqrt{11}-34}{7}$

   Nothing cancels further: $17$ and $7$ are coprime, so the fraction is in lowest terms.

5. 5

   Check the value numerically. With $\sqrt{11}=3.3166247903554$, the original is $17/5.3166247903554=3.1975173480059\,7$, and the simplified form is $(17\cdot 3.3166247903554-34)/7=22.3826214/7=3.1975173480059\,7$. The two agree to 13 decimal places, so the rationalization is correct.