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 \(\sqrt{10}+2\sqrt{2} < 6\).
