Surds and Radicals

Rationalizing a denominator

To simplify a fraction with a surd in the denominator, multiply top and bottom by the conjugate: this uses the difference-of-squares identity (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2-b^2 to clear the square root from the bottom.

Worked example

If x=2+3x = 2+\sqrt{3}, find x2+1x2x^2 + \dfrac{1}{x^2}.

Solution: Rationalize 1x\frac{1}{x} first:

1x=12+3×2−32−3=2−34−3=2−3\dfrac{1}{x} = \dfrac{1}{2+\sqrt{3}} \times \dfrac{2-\sqrt{3}}{2-\sqrt{3}} = \dfrac{2-\sqrt{3}}{4-3} = 2-\sqrt{3}

So x+1x=(2+3)+(2−3)=4x + \frac{1}{x} = (2+\sqrt{3})+(2-\sqrt{3}) = 4, and squaring both sides:

(x+1x)2=x2+1x2+2=16  ⇒  x2+1x2=14\left(x+\dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2 = 16 \;\Rightarrow\; x^2+\dfrac{1}{x^2} = 14

Worked example: a telescoping sum

Show that 13−8−18−7+17−6−16−5+15−2=5\dfrac{1}{3-\sqrt{8}} - \dfrac{1}{\sqrt{8}-\sqrt{7}} + \dfrac{1}{\sqrt{7}-\sqrt{6}} - \dfrac{1}{\sqrt{6}-\sqrt{5}} + \dfrac{1}{\sqrt{5}-2} = 5.

Solution: Every denominator here is a difference of consecutive square roots, and each has the form (a−b)(a+b)=a2−b2(a-b)(a+b)=a^2-b^2, always equal to exactly 1 once rationalized (e.g. 32−(8)2=9−8=13^2 - (\sqrt{8})^2 = 9-8=1). Rationalizing each term:

(3+8)−(8+7)+(7+6)−(6+5)+(5+2)(3+\sqrt{8}) - (\sqrt{8}+\sqrt{7}) + (\sqrt{7}+\sqrt{6}) - (\sqrt{6}+\sqrt{5}) + (\sqrt{5}+2)

Every surd term cancels with its neighbor, leaving only 3+2=53+2=5.

Continue learning

Rationalizing a denominator is the same tool used to prove that expressions like 12\frac{1}{\sqrt{2}} are irrational.