Multiply by a suitable root or conjugate without changing the value.
Extended · Syllabus E1.18
Notes and key methods
To rationalise \(\displaystyle \frac{b}{\sqrt{d}}\), multiply numerator and denominator by \(\displaystyle \sqrt{d}\). For \(\displaystyle \frac{1}{a+ \sqrt{d}}\), use the conjugate \(\displaystyle a-\sqrt{d}\) and the identity \(\displaystyle (a+\sqrt{d})(a-\sqrt{d})=a^{2}-d\).
Multiply both numerator and denominator by the same nonzero expression. State excluded zero denominators and simplify the whole result. A conjugate changes the sign between the terms; it does not negate the entire expression.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Rationalising a denominator
Let \(\displaystyle a=3, d=6\). Write \(\displaystyle \frac{1}{a+\sqrt d}=A+B\sqrt d\), where A and B are rational.
Find the rational product \(\displaystyle (a+\sqrt{d})(a-\sqrt{d})\).
Write \(\displaystyle \frac{5}{\sqrt{d}}\) as \(\displaystyle K\sqrt{d}\). Find K.
Show the reasoning
(a)
\(\displaystyle \frac{1}{a+\sqrt d}=\frac{a-\sqrt d}{a^2-d}\); here the denominator is \(3\), so the exact simplified result is \(\displaystyle 1 - \frac{\sqrt{6}}{3}\).
Answer: \(\displaystyle 1 - \frac{\sqrt{6}}{3}\)
(b)
Multiply by the appropriate root or conjugate, retain the complete denominator and simplify.
\(\displaystyle - \frac{1}{3}\)
Answer: \(\displaystyle - \frac{1}{3}\).
(c)
Multiply by the appropriate root or conjugate, retain the complete denominator and simplify.
\(3\)
Answer: \(3\).
(d)
Multiply by the appropriate root or conjugate, retain the complete denominator and simplify.
\(\displaystyle \frac{5}{6}\)
Answer: \(\displaystyle \frac{5}{6}\).
Example 2 · Rationalising a denominator
Let \(\displaystyle a=4, d=2\). Write \(\displaystyle \frac{1}{a+\sqrt d}=A+B\sqrt d\), where A and B are rational.
Find the rational product \(\displaystyle (a+\sqrt{d})(a-\sqrt{d})\).
Write \(\displaystyle \frac{6}{\sqrt{d}}\) as \(\displaystyle K\sqrt{d}\). Find K.
Show the reasoning
(a)
\(\displaystyle \frac{1}{a+\sqrt d}=\frac{a-\sqrt d}{a^2-d}\); here the denominator is \(14\), so the exact simplified result is \(\displaystyle \frac{2}{7} - \frac{\sqrt{2}}{14}\).
Find the rational product \(\displaystyle (a+\sqrt{d})(a-\sqrt{d})\).
Write \(\displaystyle \frac{7}{\sqrt{d}}\) as \(\displaystyle K\sqrt{d}\). Find K.
Show the reasoning
(a)
\(\displaystyle \frac{1}{a+\sqrt d}=\frac{a-\sqrt d}{a^2-d}\); here the denominator is \(22\), so the exact simplified result is \(\displaystyle \frac{5}{22} - \frac{\sqrt{3}}{22}\).