Review question

# Given five numbers of the form $a-b\sqrt{c}$, which is smallest? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9229

## Solution

Which of the following is smallest?

A)$10 - 3\sqrt{11} \qquad$ B) $8-3\sqrt{7} \qquad$ C) $5-2\sqrt{6} \qquad$ D) $9-4\sqrt{5} \qquad$ E) $7-4\sqrt{3}$

We could note that $10 - 3\sqrt{11} = 10-\sqrt{3^2\times 11} = \sqrt{100} - \sqrt{99}$.

Similarly, \begin{align*} 8-3\sqrt{7} &= \sqrt{64} - \sqrt{63}, \\ 5-2\sqrt{6} &= \sqrt{25}- \sqrt{24}, \\ 9-4\sqrt{5} &= \sqrt{81}-\sqrt{80}, \\ 7-4\sqrt{3} &= \sqrt{49}-\sqrt{48}. \end{align*}

So all the options are of the form $\sqrt{x} - \sqrt{x-1}$.

What does the graph of $y = \sqrt{x} - \sqrt{x-1}$ look like? It seems intuitively obvious that as $x$ gets bigger, $y$ gets smaller.

For example, $\sqrt{100}-\sqrt{99}$ is surely bigger than $\sqrt{10\,000} - \sqrt{9\,999}$.

So we need to pick the largest value of $x$ on offer, and the answer is (A).

We could note that $(\sqrt{x} - \sqrt{x-1} )(\sqrt{x} + \sqrt{x-1} ) =1$, and so $\sqrt{x} - \sqrt{x-1} = \dfrac{1}{\sqrt{x} + \sqrt{x-1}}$, which certainly gets smaller as $x$ increases.