Multiplying and dividing surds

Multiplying and dividing surds

What is √2 · √8? A calculator shows 1.414… · 2.828… = 4. Two numbers with never-ending decimals gave a whole number. That's no accident: √2 · √8 = √16 = 4.

When you multiply, surds are much friendlier than when you add. You can always multiply them, even when the numbers under the roots are different. (In the US, surds are called radicals.)

Contents

Multiplying surds

√a · √b = √(a · b) (for a ≥ 0, b ≥ 0)

To multiply two surds, multiply the numbers under them. Then simplify the answer.

Worked example 1

√10 · √2

  1. √10 · √2 = √20.
  2. √20 = √(4 · 5) = 2√5.

Worked example 2

√3 · √12

  1. √3 · √12 = √36 = 6.

A surd multiplied by itself gives the number under it: √a · √a = a. For example, √7 · √7 = 7.

Dividing surds

√a ÷ √b = √(a ÷ b) (for a ≥ 0, b > 0)

Worked example 3

√80 ÷ √5

  1. √80 ÷ √5 = √(80 ÷ 5) = √16 = 4.

Worked example 4

√54 ÷ √3

  1. √(54 ÷ 3) = √18 = 3√2.

Numbers in front of the root

Multiply the numbers in front of the roots together, and the roots together.

Worked example 5

2√6 · 3√10

  1. 2 · 3 = 6 and √6 · √10 = √60.
  2. √60 = √(4 · 15) = 2√15.
  3. 6 · 2√15 = 12√15.

Expanding a bracket

Multiply the surd in front of the bracket by every term inside the bracket, just like with expressions: a(b + c) = ab + ac.

Worked example 6

√5 · (√20 − 3)

  1. √5 · √20 = √100 = 10.
  2. √5 · 3 = 3√5.
  3. The answer: 10 − 3√5.

The answer has two parts: a number without a root (10) and a multiple of a root (−3√5). They cannot be combined, just like 10 apples and 3 pears. We write an answer like this in the form p + q√r.

Identities with surds

The familiar identities still work:

  • (a + b)(a − b) = a² − b²
  • (a + b)² = a² + 2ab + b²
  • (a − b)² = a² − 2ab + b²

Worked example 7

(√3 + 5)(√3 − 5)

  1. Using the identity: (√3)² − 5².
  2. (√3)² = 3 and 5² = 25.
  3. 3 − 25 = −22.

The root has disappeared. We will use this trick when rationalizing the denominator.

Worked example 8

(√2 + √3)²

  1. a = √2, b = √3: a² = 2, b² = 3, 2ab = 2√6.
  2. 2 + 2√6 + 3 = 5 + 2√6.

Worked example 9

(2 + √3)(1 − √3)

Multiply every term by every term:

  1. 2 · 1 = 2
  2. 2 · (−√3) = −2√3
  3. √3 · 1 = √3
  4. √3 · (−√3) = −3
  5. Altogether: 2 − 2√3 + √3 − 3 = −1 − √3.

Common mistakes

  • (√2 + √3)² = 2 + 3. The middle term 2ab = 2√6 is missing. The correct answer is 5 + 2√6.
  • 2√3 · 4√3 = 8√3. The roots must be multiplied too: √3 · √3 = 3, so 2√3 · 4√3 = 8 · 3 = 24.
  • Not simplifying the answer. √10 · √2 = √20 must be finished as 2√5.
  • Writing (√a)² as 2√a. Squaring a square root gives the number under it: (√7)² = 7.

Summary

  • √a · √b = √(ab), √a ÷ √b = √(a ÷ b), √a · √a = a.
  • Multiply the numbers in front of the roots separately from the roots, then simplify the answer.
  • Expand brackets as you would with expressions, or use the identities.
  • Write the answer in the form p + q√r.

Try it yourself

  1. √6 · √15
  2. √98 ÷ √2
  3. √2 · (√18 + 4)
  4. (√7 − 2)(√7 + 2)
  5. (√5 − 1)²
Answers:
  1. √90 = √(9 · 10) = 3√10.
  2. √49 = 7.
  3. √36 + 4√2 = 6 + 4√2.
  4. 7 − 4 = 3.
  5. 5 − 2√5 + 1 = 6 − 2√5.

Practice

👉 Next: Rationalizing the denominator