Stepwise
Start practisingPractise

Surds · Lesson 2 of 2

Rationalising the denominator

Removing surds from the bottom of a fraction: multiply by the surd for a single root, and by the conjugate for a sum or difference.

14 minYou should already know: Indices & standard form
  1. 1
  2. 2

An answer in surd form shouldn’t have a surd in the denominator. Removing it is called rationalising. You multiply the top and bottom by the same thing, so the value doesn’t change, only its form. You need simplifying surds first.

One surd on the bottom

Multiply top and bottom by that surd, because a×a=a\sqrt a \times \sqrt a = a:

6√3×√3√3=6√33=2√3this is 1
Multiply by 1√3 ÷ √3 is 1, so the value doesn't change

More: one surd on the bottom

A sum or difference on the bottom: the conjugate

For 1a+b\frac{1}{a + \sqrt b}, multiplying by b\sqrt b doesn’t clear the surd. Instead multiply by the conjugate, the same two terms with the sign between them changed. The difference of two squares removes the surd:

(a+b)(a−b)=a2−b(a + \sqrt b)(a - \sqrt b) = a^2 - b

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q3 (a)

Without using mathematical tables or calculators, simplify 332−423−243\sqrt{\frac32} - 4\sqrt{\frac23} - \sqrt{24}.

  1. Rationalise each surd

    332=332×22=3623\sqrt{\frac32} = \frac{3\sqrt3}{\sqrt2} \times \frac{\sqrt2}{\sqrt2} = \frac{3\sqrt6}{2}, and 423=423×33=4634\sqrt{\frac23} = \frac{4\sqrt2}{\sqrt3} \times \frac{\sqrt3}{\sqrt3} = \frac{4\sqrt6}{3}.

    Think first. 32=32\sqrt{\frac32} = \frac{\sqrt3}{\sqrt2}. What do you multiply by?

  2. The third term

    24=4×6=26\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt6.

  3. Collect over a common denominator

    966−866−1266=−1166\frac{9\sqrt6}{6} - \frac{8\sqrt6}{6} - \frac{12\sqrt6}{6} = -\frac{11\sqrt6}{6}

    Think first. What common denominator do 2 and 3 need?

More: rationalising with the conjugate

Matching a form x + y√n

Some questions give the answer’s form and ask for xx and yy. Rationalise, then compare the whole-number parts and the surd parts.

More: matching a form

Your turn

JAMB 1997 · UME · Q7

Simplify 23+3535−23\dfrac{2\sqrt3 + 3\sqrt5}{3\sqrt5 - 2\sqrt3}.

Worked solution (try it first)
  1. Multiply the top and bottom by the conjugate of the bottom, 35+233\sqrt5 + 2\sqrt3.
  2. Bottom: (35)2−(23)2=45−12(3\sqrt5)^2 - (2\sqrt3)^2 = 45 - 12, which is 33.
  3. Top: (23+35)2=12+1215+45(2\sqrt3 + 3\sqrt5)^2 = 12 + 12\sqrt{15} + 45, which is 57+121557 + 12\sqrt{15}.
  4. Divide top and bottom by 3: 19+41511\dfrac{19 + 4\sqrt{15}}{11}, option A.

Report a problem with this question

More past questions like this