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

## Suggestion

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) $\dfrac{5}{2}$

Could we write $2016$ as the product of its prime factors? Could we do the same for $56$?

How could we now simplify the square roots?