Multiplying and dividing surds — Mathematics

Before this

Simplifying surds
Adding and subtracting surds
Expanding brackets · coming soon

This leads to

Rationalising the denominator
Difference of two squares
The quadratic formula

Two irrational sides, one whole-number area

A rectangle is $3$ cm wide and $12$ cm long. Both sides are irrational. Their decimals never end and never repeat. So what is the area?

Multiply on a calculator: $1.7320508 \dots \times 3.4641016 \dots$ . The answer on the screen is 6. Not roughly 6. Exactly 6.

In the last lecture, surds were stubborn. $2 + 3$ refused to become a single surd, and we proved it never could. Multiplication behaves completely differently. Two surds can multiply to give a whole number. This lecture shows why, and then uses it to multiply out brackets that contain surds. By the end, you will see two irrational numbers whose product is a whole number every time, by design.

Multiplying: the product rule again

You have already proved the rule that does the work. In the lecture on simplifying surds, you saw that for numbers $a$ and $b$ that are zero or positive:

a \times b = ab

The reason was that $a \times b$ is positive and squares to $ab$ , which is exactly what $ab$ means. Back then, we used the rule to split a root apart: $20 = 4 \times 5$ . Now we use it in the other direction, to join two roots together.

For the rectangle: $3 \times 12 = 36 = 6$ . The product inside the root, 36, happens to be a perfect square. That is the whole secret of the hook.

When there are numbers in front, multiplication lets you change the order. So multiply the numbers in front together, and the roots together:

32 \times 45 = (3 \times 4) \times (2 \times 5) = 1210

One special case deserves its own line, because it appears in almost every calculation with brackets. A square root times itself gives back the number inside:

a \times a = a (for a \geq 0)

This is not a new rule. It is the definition of a square root: $7$ is the positive number that multiplies by itself to give 7. So $7 \times 7 = 7$ . A surd multiplied by itself is no longer a surd.

Dividing: the same rule, upside down

Division follows from the same idea. For $a \geq 0$ and $b > 0$ :

Multiplying and dividing surds

$a \times b = ab$ and $\frac{a}{b} = \frac{a}{b}$

Numbers in front of the roots multiply or divide separately. A root times itself gives the number inside: $a \times a = a$ .

Why does the division rule hold? Square the left side: $(\frac{a}{b})^{2} = \frac{a \times a}{b \times b} = \frac{a}{b}$ . So $\frac{a}{b}$ is a positive number that squares to $\frac{a}{b}$ . That is exactly the definition of $\frac{a}{b}$ .

For example, $18 \div 2 = 18 \div 2 = 9 = 3$ . Another irrational pair gives a whole number.

Stop and think

A student says: "An irrational number times an irrational number must be irrational." Is that true?

Show the answer

No. $2 \times 8 = 16 = 4$ , and $3 \times 12 = 6$ . Both factors are irrational, but the products are whole numbers. Sometimes the product is irrational, as in $2 \times 3 = 6$ . It depends on whether the product inside the root is a perfect square. So "irrational times irrational" can be either kind of number. You must check each case.

Brackets: every term meets every term

Now the main event. How do you multiply $(2 + 3) (1 + 5)$ ?

Think of it as the area of a rectangle with sides $2 + 3$ and $1 + 5$ . Cut the rectangle along the joins between the parts. You get four smaller rectangles, and the whole area is their total.

The four parts of the rectangle are the four products. Each term in one bracket meets each term in the other.

Each small rectangle is one term times one term:

(2 + 3) (1 + 5) = 2 + 25 + 3 + 15

Can anything be collected? The terms are a whole number and three surds with radicands 5, 3 and 15. None of these are like surds, so this is the final answer. Check with decimals: $(2 + 1.732) (1 + 2.236) = 3.732 \times 3.236 \approx 12.08$ , and $2 + 4.472 + 1.732 + 3.873 \approx 12.08$ .

Some books call this method FOIL: First, Outer, Inner, Last. The name is only a checklist for the same four rectangles. Whatever you call it, the rule is the same: every term in the first bracket multiplies every term in the second.

Signs need care, so step through a harder example below, one move at a time.

Step through — expanding (√5 + 2)(√5 − 3)

Press Next to add one step. Read the reason before you go on.

Each step gives one line and the reason it is allowed.

(√5 + 2)(√5 − 3): four products are needed, one for each pair of terms.
√5 × √5 = 5: a square root times itself gives the number inside.
√5 × (−3) = −3√5: the minus sign belongs to the 3 and comes with it.
2 × √5 = 2√5.
2 × (−3) = −6: a positive times a negative is negative.
5 − 3√5 + 2√5 − 6: all four products written down before anything is collected.
(5 − 6) + (−3√5 + 2√5): whole numbers together, √5 terms together.
= −1 − √5: −3√5 + 2√5 = −1√5, which is written −√5.

Result: (√5 + 2)(√5 − 3) = −1 − √5 ≈ −3.24. Decimal check: 4.236 × (−0.764) ≈ −3.24.

Squaring a bracket: do not lose the middle

Squaring a bracket means multiplying it by itself. So $(3 + 2)^{2} = (3 + 2) (3 + 2)$ , and the four products are:

9 + 32 + 32 + 2 = 11 + 62

A very common wrong answer is $3^{2} + (2)^{2} = 11$ . It squares each term and forgets the two middle rectangles. Those two rectangles are worth $62 \approx 8.49$ , which is far too much to lose. Check: $(3 + 1.414)^{2} = 4.41 4^{2} \approx 19.49$ , and $11 + 8.49 = 19.49$ . The answer $11$ alone fails the check badly.

Conjugates: when the surds vanish completely

Now look at a very special pair of brackets: $(7 + 2) (7 - 2)$ . The two brackets are the same except for one sign. Expand them:

7 \times 7 - 27 + 27 - 4 = 7 - 4 = 3

The two middle terms are equal and opposite, so they cancel. The first term is $7 \times 7 = 7$ , a whole number. So the answer is exactly 3. Every surd has vanished.

That is it. That is the whole idea. Two brackets like this, $a + b$ and $a - b$ , are called a conjugate pair. Their product is always a whole number when $a$ and $b$ are whole numbers:

(a + b) (a - b) = a^{2} - a b + a b - b = a^{2} - b

Try any pair you like below, and watch the middle terms cancel every time.

Explore — conjugate pairs always give a whole number

Choose a and b. Watch the four products, the two middle terms cancelling, and the decimal check.

a = 2

b = 7

Some conjugate pairs and their products. In every case the middle terms cancel and the answer is a² − b.

Pair	a² − b	Decimal check
(2 + √7)(2 − √7)	4 − 7 = −3	4.646 × (−0.646) ≈ −3.00
(3 + √7)(3 − √7)	9 − 7 = 2	5.646 × 0.354 ≈ 2.00
(2 + √3)(2 − √3)	4 − 3 = 1	3.732 × 0.268 ≈ 1.00
(4 + √11)(4 − √11)	16 − 11 = 5	7.317 × 0.683 ≈ 5.00

Notice what you can now do. Given any expression $a + b$ , you know a partner that multiplies with it to give a whole number. The next lecture uses exactly this trick to remove surds from the bottom of a fraction.

Worked examples

Worked Example 1 — Multiply, then simplify

Simplify $6 \times 10$ .

Join the roots: $6 \times 10 = 60$ . Now simplify, as in the earlier lecture. The square factors of 60 are 4 only (9, 16, 25 and 36 do not divide 60). So

60 = 4 \times 15 = 215

Check: $6 \times 10 \approx 2.449 \times 3.162 \approx 7.746$ , and $215 \approx 2 \times 3.873 = 7.746$ .

Worked Example 2 — Dividing with numbers in front

Simplify $615 \div 23$ .

Divide the numbers in front, and divide the roots:

\frac{6 15}{2 3} = \frac{6}{2} \times \frac{15}{3} = 35

Check by multiplying back: $23 \times 35 = 615$ . Correct.

Worked Example 3 — One term times a bracket

Expand $3 (3 + 4)$ .

The $3$ outside multiplies each term inside: $3 \times 3 = 3$ and $3 \times 4 = 43$ . So the answer is $3 + 43$ .

Check: $1.732 \times (1.732 + 4) = 1.732 \times 5.732 \approx 9.93$ , and $3 + 6.93 = 9.93$ .

Worked Example 4 — An exam-style area

A rectangle has sides $(6 + 2)$ cm and $(6 - 2)$ cm. Show that its area is a whole number.

This is a conjugate pair, with $6$ in place of a whole number. The same cancelling happens:

(6 + 2) (6 - 2) = 6 - 12 + 12 - 2 = 4

The area is exactly 4 cm². Check: $3.864 \times 1.035 \approx 4.00$ .

Three classic mistakes

Trap 1 — Mixing up adding and multiplying

$2 \times 2 = 2$ , but $2 + 2 = 22$ . Students often swap these, or write $2 \times 2 = 4$ . Multiplying joins the roots. Adding only collects like surds.

Trap 2 — Squaring a bracket term by term

$(3 + 2)^{2}$ is not $9 + 2$ . It is $11 + 62$ . A squared bracket is two brackets multiplied, so it has two middle terms. Draw the four rectangles if you are unsure.

Trap 3 — Mixing the numbers in front with the roots

$32 \times 45 = 1210$ . Writing $710$ adds the numbers in front instead of multiplying them. Writing $127$ adds the numbers inside the roots. Numbers in front multiply numbers in front, and roots multiply roots.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Simplify $26 \times 32$ and $45 \div 5$ .

Hint 1
Multiply the numbers in front, then join the roots. Simplify the root at the end.

Hint 2
$2 \times 3 = 6$ and $6 \times 2 = 12$ . For the division, $45 \div 5 = 9$ .

Full answer
$26 \times 32 = 612 = 6 \times 23 = 123$ . $45 \div 5 = 9 = 3$ . Check: $123 \approx 20.78$ , and $4.899 \times 4.243 \approx 20.78$ .
Expand and simplify $(4 - 3) (2 + 3)$ .

Hint 1
Four products: $4 \times 2$ , $4 \times 3$ , $- 3 \times 2$ , $- 3 \times 3$ .

Hint 2
The last product is $- 3$ . Then collect the whole numbers and the $3$ terms.

Full answer
$8 + 43 - 23 - 3 = 5 + 23$ . Check: $2.268 \times 3.732 \approx 8.46$ , and $5 + 3.46 = 8.46$ .
Expand $(25 - 1)^{2}$ .

Hint 1
Write it as $(25 - 1) (25 - 1)$ . There are two middle terms.

Hint 2
$25 \times 25 = 4 \times 5 = 20$ .

Full answer
$20 - 25 - 25 + 1 = 21 - 45$ . Check: $(4.472 - 1)^{2} = 3.47 2^{2} \approx 12.06$ , and $21 - 8.94 = 12.06$ .
The rectangle in Worked Example 4 has sides $(6 + 2)$ cm and $(6 - 2)$ cm. Find the exact length of its diagonal.

Hint 1
Use Pythagoras' theorem: diagonal² = side² + side². Square each bracket first.

Hint 2
$(6 + 2)^{2} = 8 + 212$ and $(6 - 2)^{2} = 8 - 212$ . Add them.

Full answer
$(6 + 2)^{2} = 6 + 212 + 2 = 8 + 212$ . $(6 - 2)^{2} = 8 - 212$ . Adding, the surds cancel: diagonal² $= 16$ . So the diagonal is exactly 4 cm. Check: $3.86 4^{2} + 1.03 5^{2} \approx 14.93 + 1.07 = 16.00$ . A rectangle whose sides are both irrational has a whole-number area and a whole-number diagonal.

The bigger idea

Adding surds keeps you among the surds. Multiplying them can take you back to whole numbers. That difference is why surds are so useful in exact work: with the right partner, an irrational number can be turned into a whole number on purpose.

Mathematicians use this again and again. The quadratic formula produces answers like $3 + 5$ and $3 - 5$ in conjugate pairs, and their product has no surd left in it. In the next lecture, the same idea cleans the surds out of the bottom of a fraction.

Key concepts in this lesson

For numbers that are zero or positive, $a \times b = ab$ , and for $b > 0$ , $a \div b = a \div b$ .
A square root multiplied by itself gives the number inside, so $7 \times 7 = 7$ .
When multiplying surds with numbers in front, the numbers in front multiply together and the roots multiply together.
To expand brackets, every term in the first bracket multiplies every term in the second, which is the area of four rectangles.
A conjugate pair $(a + b) (a - b)$ multiplies to $a^{2} - b$ , because the two middle terms cancel.