Year 10 Extended · Unit 1 Number sets, surds, indices and bounds · Lesson 3 of 6

When does a product of two irrational numbers come out whole?

Brackets with surds expand by the same rules as algebra. This lesson asks when a product of two irrational numbers can come out whole, and how that clears a denominator with two terms.

Number and AlgebraNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. Simplify √12.
    Answer2√3. 12 = 4 × 3 and √4 = 2.
  2. Expand (x + 3)(x − 3).
    Answerx² − 9. The middle terms −3x and +3x cancel.
  3. From Lesson 1: is √2 × √8 rational or irrational?
    AnswerRational. √2 × √8 = √16 = 4.
Words used in this lesson
surd
a root that cannot be written as a whole number or a fraction. For example: √11.
conjugate
the same two terms with the sign between them switched. For example: 3 + √5 and 3 − √5.
rationalize
rewrite a fraction so the denominator is a whole number. For example: 1/√3 = √3/3.
like surds
surds with the same root, which can be added or subtracted. For example: 4√3 and 9√3.
cross term
the middle term when a sum is squared. For example: the 2ab in (a + b)².
difference of two squares
a product that gives one square minus another. For example: (x + 4)(x − 4).

A tile with two irrational edges

A craftsperson in Isfahan cuts a rectangular tile that is (5 + √7) cm by (5 − √7) cm. Both edges are irrational lengths that cannot be measured exactly, so it seems the area must be irrational too. Yet the craftsperson can say the area exactly.

Predict first: is that area rational or irrational? And is (√2 + √3)² equal to 5, or to something that still has a root in it? Use the two figures to decide.

4
3
13area
(4 + √3)(4 − √3)rectangle

(4 + √3)(4 − √3) = 4² − 3 = 13, a whole number.

Show me the steps
  1. a = 4 and b = 3: the square has area 16, the cut-out has area 3, so (4 + √3)(4 − √3) = 16 − 3 = 13.
  2. Slide a to 6 with b = 3: (6 + √3)(6 − √3) = 36 − 3 = 33. The result is a whole number again.
  3. Set a = 3 and b = 8: 9 − 8 = 1. The rectangle is 3 + √8 = 5.83… by 3 − √8 = 0.17… and is very thin, but its area is exactly 1.
  4. Whatever a and b you choose, the two irrational parts cancel: the middle terms −a√b and +a√b add to zero, and (√b)² = b.
3
5
8 + 2√15total area
√15each rectangle

(√3 + √5)² = 3 + 5 + 2√15 = 8 + 2√15.

Show me the steps
  1. p = 3 and q = 5: the parts are 3, 5 and two rectangles of √15 each, so the area is 8 + 2√15. It is not 8.
  2. Set p = 4 and q = 9: the sides are 2 and 3, the big square has side 5 and area 25, and the formula gives 13 + 2√36 = 13 + 12 = 25.
  3. Set p = 2 and q = 8: √16 = 4, so the total is 10 + 8 = 18, a whole number, because √2 + √8 = 3√2 and (3√2)² = 18.
  4. Set p = q = 3: the rectangles are squares of area 3, so the total is 6 + 6 = 12, which is (2√3)².

Conjugates multiply to a rational number. (a + √b)(a − √b) = a² − b, because the middle terms cancel and (√b)² = b. Two brackets with surds are expanded like any brackets, then like surds are collected.

To rationalize a two-term denominator, multiply the top and the bottom by the conjugate of the bottom. The bottom becomes a² − b, a whole number.

A square of a sum has a cross term: (√p + √q)² = p + q + 2√(pq).

Two common mistakes. Writing (1 + √3)² as 1 + 3 = 4: the cross term is missing, and it is 4 + 2√3. And thinking √(a + b) = √a + √b: √(9 + 16) = 5, but √9 + √16 = 7.

Worked examples

1. Expand and simplify (2√3 + 1)(2√3 − 1).

  1. This is a conjugate pair with a = 2√3 and 1: the middle terms cancel.
  2. (2√3)² − 1² = 4 × 3 − 1 = 12 − 1 = 11.

2. Rationalize 10/(√6 − 1).

  1. The conjugate of √6 − 1 is √6 + 1. Multiply the top and the bottom by it.
  2. Bottom: (√6 − 1)(√6 + 1) = 6 − 1 = 5. Top: 10(√6 + 1).
  3. 10(√6 + 1)/5 = 2√6 + 2.

3. Expand (3 + √2)².

  1. (3 + √2)(3 + √2) = 9 + 3√2 + 3√2 + 2.
  2. Collect: 11 + 6√2. The cross term 2 × 3 × √2 is the 6√2, and it does not vanish: 11 + 6√2.

Your go. The first step is given. Rationalize 8/(3 − √5).

  1. Multiply the top and the bottom by the conjugate, 3 + √5.
  2. Work out the new bottom, then simplify the fraction.
Check the answer

Bottom: 9 − 5 = 4. Top: 8(3 + √5). So 8(3 + √5)/4 = 2(3 + √5) = 6 + 2√5.

Check before you go on

1. Work out (7 − √3)(7 + √3).

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. √50 + √18 − √8 can be written as a√2. What is a?

3. (2 + √3)(5 − √3) can be written as a + b√3. What is b?

Level 2 · Building

4. 24/(√7 − √3) can be written as a√7 + a√3. What is a?

5. Work out (√5 + √20)².

6. A square tile has sides of length (√5 + 2) cm. Its area is a + b√5 cm². What is b?

Level 3 · Stretch

7. x = 3 + √5. Find the exact value of x² − 6x.

8. √(17 + 12√2) = 3 + p√2, where p is a whole number. Find p. (Square both sides.)

Something to think about

Rationalizing 1/(√7 − √3) multiplies by the conjugate. Would multiplying top and bottom by √7 − √3 again (the same, not the conjugate) have helped? Try it, and say what is special about the conjugate.

One way to think about it

(√7 − √3)² = 7 − 2√21 + 3 = 10 − 2√21, which still has a root, so the denominator is not cleared. Only the conjugate √7 + √3 makes the middle terms cancel, giving (√7)² − (√3)² = 4. That is why a conjugate is the same two terms with the sign between them switched.

Exit ticket

9. Write 12/(4 − √10) with a whole-number denominator, as a + b√10 in simplest form.