Review question

# Can we show this surd is less than $6$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5701

## Solution

Write down the square of $\sqrt{a}+b\sqrt{c}$ in the form $p+q\sqrt{r}$.

We have $(\sqrt{a}+b\sqrt{c})^2 = (\sqrt{a}+b\sqrt{c})(\sqrt{a}+b\sqrt{c}) = a+2b\sqrt{ac}+b^2c.$

Comparing this to the stated form in the question reveals $p = a + b^2c, q = 2b, r = ac.$

Without evaluating square roots or using tables [or calculators], show that $\sqrt{10}+2\sqrt{2}$ is less than $6$.

To show $\sqrt{10}+2\sqrt{2} < 6$, we will want to use the result from the first part, which will mean squaring both sides.

We should always be careful when squaring an inequality!

We can write $-2<1$, which is true, but if we square both sides, we get $4<1$, which is untrue.

BUT… if $0 < a < b$, then we CAN say that $0 < a^2 < b^2$.

We have that $\left( \sqrt{10}+2\sqrt{2} \right)^2 = 10+2^2\times2 + 2\times2 \sqrt{10 \times 2} = 18 + 4 \sqrt{20},$ so

\begin{align*} \sqrt{10}+2\sqrt{2} &< 6\\ \iff 18 + 4 \sqrt{20} &< 36\\ \iff 4 \sqrt{20} &< 18\\ \iff \sqrt{20} &< \frac{9}{2}\\ \iff 20 &< \frac{81}{4}\\ \iff 80 &< 81. \end{align*}

This is true, so we can say with certainty that $\sqrt{10}+2\sqrt{2} < 6$.