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Worked solution

Recall rationalisation before complex division

How do we rationalize 1a+b\dfrac{1}{\sqrt a+\sqrt b}?

CJC · Division opening activity

Answer

a−ba−b\dfrac{\sqrt a-\sqrt b}{a-b}, for real a,b≥0a,b\ge0 with a≠ba\ne b.

Compare your reasoning

Multiply by the conjugate of the surd denominator:

1a+b=1a+b⋅a−ba−b=a−ba−b.\begin{aligned}\frac1{\sqrt a+\sqrt b}&=\frac1{\sqrt a+\sqrt b}\cdot\frac{\sqrt a-\sqrt b}{\sqrt a-\sqrt b}\\&=\frac{\sqrt a-\sqrt b}{a-b}.\end{aligned}

For example, take a=5a=5, b=1b=1. Then

15+1=5−14=54−14.\frac1{\sqrt5+1}=\frac{\sqrt5-1}{4}=\frac{\sqrt5}{4}-\frac14.

This makes the denominator real and, in this example, an integer, so the fraction splits into two terms. Complex division uses the same idea with the denominator’s complex conjugate.

How did this question go?

Common mistakes

The source calls the new denominator an integer; in the general formula it is a−ba-b, which need not be an integer without an assumption on a,ba,b. The displayed transformation requires a≠ba\ne b. When a=b>0a=b>0, the original fraction still exists and is 1/(2a)1/(2\sqrt a); multiplying by the zero conjugate would be invalid. When a=b=0a=b=0, the original denominator is zero.