Review question

# Can we write $\sqrt{2016}+\sqrt{56}$ as a power of $14$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7658

## Solution

For which value of $k$ is $\sqrt{2016}+\sqrt{56}$ equal to $14^k$?

A)$\:\dfrac{1}{2} \qquad$ B) $\dfrac{3}{4} \qquad$ C) $\dfrac{5}{4} \qquad$ D) $\dfrac{3}{2} \qquad$ E) $\dfrac{5}{2}$

If we factorise $2016$ into its prime factors, we find $2016 = 2^5\times 3^2 \times 7$.

Doing the same to $56$, we find $56 = 2^3\times 7$. Thus

\begin{align*} \sqrt{2016}+\sqrt{56} &= \sqrt{2^5\times 3^2 \times 7}+\sqrt{2^3\times 7} \\ &= \sqrt{(2^4\times 3^2) \times 14}+\sqrt{2^2\times 14}\\ &= 2^2 \times 3 \times \sqrt{14}+2\sqrt{14}= 14\sqrt{14}\\ &= 14^{3/2}, \end{align*}

so $k = \dfrac{3}{2}$ and the answer is (D).