Square roots with multiplication and division

A rectangular garden is $\sqrt{50}$ m long and $\sqrt8$ m wide. Its area is $\sqrt{50}\cdot\sqrt8$ m². A calculator gives $7.07\ldots\cdot2.82\ldots$ , which is a messy multiplication. Is there a way to find the exact area without a calculator?

The rectangular garden

The square root of a product

Explore: compare $\sqrt{ab}$ with $\sqrt a\cdot\sqrt b$

We work out both for a few pairs of perfect squares:

$a$ , $b$	$\sqrt{ab}$	$\sqrt a\cdot\sqrt b$
$4$ , $9$	$\sqrt{36}=6$	$2\cdot3=6$
$16$ , $25$	$\sqrt{400}=20$	$4\cdot5=20$
$1$ , $49$	$\sqrt{49}=7$	$1\cdot7=7$

Both ways always give the same result.

Why? The number $\sqrt a\cdot\sqrt b$ is non-negative, and $\left(\sqrt a\cdot\sqrt b\right)^2=\left(\sqrt a\right)^2\cdot\left(\sqrt b\right)^2=ab$ . So $\sqrt a\cdot\sqrt b$ is the principal square root of $ab$ .

Reading the rule from left to right, we take the square root of a product: take the root of each factor, then multiply.

Reading the rule from right to left, we multiply square roots: multiply the numbers under the root signs, then take the root of the result. This answers the garden question.

The rule still holds when $a$ and $b$ are non-negative expressions: $\sqrt{AB}=\sqrt A\cdot\sqrt B$ for $A\ge0$ , $B\ge0$ .

The square root of a quotient

We have $\sqrt{\dfrac{36}{4}}=\sqrt9=3$ and $\dfrac{\sqrt{36}}{\sqrt4}=\dfrac62=3$ . Division also "passes through" the root sign, just like multiplication.

The reason is similar: $\dfrac{\sqrt a}{\sqrt b}\ge0$ and $\left(\dfrac{\sqrt a}{\sqrt b}\right)^2=\dfrac ab$ . The number $b$ must be positive because it is in the denominator.

The other way round, to divide two roots we divide the numbers under the root signs and then take the root.

Real-life applications

Many scientific formulas contain square roots. The two rules help us calculate quickly and exactly.

Key terms

English	Tiếng Việt
take the square root	khai căn bậc hai
product	tích
quotient	thương
factor	thừa số
perfect square	số chính phương
mixed number	hỗn số
square-root expression	căn thức bậc hai

Summary

$\sqrt{ab}=\sqrt a\cdot\sqrt b$ for $a\ge0$ , $b\ge0$ : take the root of each factor, or multiply roots into one root.
$\sqrt{\dfrac ab}=\dfrac{\sqrt a}{\sqrt b}$ for $a\ge0$ , $b>0$ : take the root of the numerator and the denominator, or divide roots into one root.
Both rules hold for non-negative expressions $A$ , $B$ too (with a positive denominator).
Tip: regroup the factors to get perfect squares, as in $810\cdot40=81\cdot400$ .
There is no rule like this for addition or subtraction.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Evaluate: a) $\sqrt{16\cdot0.25}$ ; b) $\sqrt{2.5\cdot40}$ ; c) $\sqrt{12\cdot75}$ .

What you need and how to start

What you need: $\sqrt{ab}=\sqrt a\cdot\sqrt b$ for $a$ , $b\ge0$ .

Steps:

If each factor is a perfect square, take the root of each factor.
If not, multiply to get a perfect square.

Show solution

a) $\sqrt{16\cdot0.25}=\sqrt{16}\cdot\sqrt{0.25}=4\cdot0.5=2$ . b) $\sqrt{2.5\cdot40}=\sqrt{100}=10$ . c) $\sqrt{12\cdot75}=\sqrt{900}=30$ .

Answer. a) $2$ ; b) $10$ ; c) $30$ .

Exercise 2

Evaluate: a) $\sqrt3\cdot\sqrt{27}$ ; b) $\sqrt{0.2}\cdot\sqrt{180}$ ; c) $\sqrt{1.6}\cdot\sqrt{250}$ .

What you need and how to start

What you need: $\sqrt a\cdot\sqrt b=\sqrt{ab}$ .

Steps:

Multiply the two numbers under the root signs.
Take the root of the result.

Show solution

a) $\sqrt3\cdot\sqrt{27}=\sqrt{81}=9$ . b) $\sqrt{0.2}\cdot\sqrt{180}=\sqrt{36}=6$ . c) $\sqrt{1.6}\cdot\sqrt{250}=\sqrt{400}=20$ .

Answer. a) $9$ ; b) $6$ ; c) $20$ .

Exercise 3

Evaluate: a) $\sqrt{\dfrac{49}{81}}$ ; b) $\sqrt{2\dfrac79}$ .

What you need and how to start

What you need: $\sqrt{\dfrac ab}=\dfrac{\sqrt a}{\sqrt b}$ for $a\ge0$ , $b>0$ .

Steps:

Change the mixed number into a fraction.
Take the root of the numerator and of the denominator.

Show solution

a) $\sqrt{\dfrac{49}{81}}=\dfrac{\sqrt{49}}{\sqrt{81}}=\dfrac79$ . b) We have $2\dfrac79=\dfrac{25}9$ , hence $\sqrt{2\dfrac79}=\dfrac{\sqrt{25}}{\sqrt9}=\dfrac53$ .

Answer. a) $\dfrac79$ ; b) $\dfrac53$ .

Exercise 4

Evaluate: a) $\dfrac{\sqrt{72}}{\sqrt2}$ ; b) $\dfrac{\sqrt{6.4}}{\sqrt{0.1}}$ ; c) $\dfrac{\sqrt{12.5}}{\sqrt{0.5}}$ .

What you need and how to start

What you need: $\dfrac{\sqrt a}{\sqrt b}=\sqrt{\dfrac ab}$ .

Steps:

Divide the two numbers under the root signs.
Take the root of the quotient.

Show solution

a) $\dfrac{\sqrt{72}}{\sqrt2}=\sqrt{36}=6$ . b) $\dfrac{\sqrt{6.4}}{\sqrt{0.1}}=\sqrt{64}=8$ . c) $\dfrac{\sqrt{12.5}}{\sqrt{0.5}}=\sqrt{25}=5$ .

Answer. a) $6$ ; b) $8$ ; c) $5$ .

Exercise 5

Simplify the following expressions: a) $\sqrt{5x}\cdot\sqrt{20x}$ for $x\ge0$ ; b) $\dfrac{\sqrt{27y^3}}{\sqrt{3y}}$ for $y>0$ .

What you need and how to start

What you need: combine into one root, then use $\sqrt{A^2}=|A|$ .

Steps:

Multiply (or divide) the expressions under the root signs.
Write the result as a square and remove the root using the condition.

Show solution

a) We have $\sqrt{5x}\cdot\sqrt{20x}=\sqrt{100x^2}=|10x|=10x$ (since $x\ge0$ ). b) We have $\dfrac{\sqrt{27y^3}}{\sqrt{3y}}=\sqrt{9y^2}=|3y|=3y$ (since $y>0$ ).

Answer. a) $10x$ ; b) $3y$ .

Exercise 6

Simplify the expression $\sqrt{4a^2}-3a$ for $a<0$ .

What you need and how to start

What you need: $\sqrt{4a^2}=\sqrt4\cdot\sqrt{a^2}=2|a|$ (see square roots and square-root expressions).

Steps:

Write $\sqrt{4a^2}=2|a|$ .
For $a<0$ , $|a|=-a$ .

Show solution

We have $\sqrt{4a^2}-3a=2|a|-3a$ . Since $a<0$ , $|a|=-a$ . Hence $\sqrt{4a^2}-3a=-2a-3a=-5a$ .

Answer. $-5a$ .

Exercise 7

A square photo has an area of $150$ cm². It is enlarged into a square with $6$ times the area.

a) How long, in centimetres, is a side of the new photo?

b) How many times as long as the old side is the new side (rounded to two decimal places)?

What you need and how to start

What you need: a square of area $S$ has side $\sqrt S$ ; $\dfrac{\sqrt{6S}}{\sqrt S}=\sqrt6$ .

Steps:

The new area is $150\cdot6$ ; take its root.
Divide the two sides using the rule for dividing roots.

Show solution

a) The area of the new photo is $150\cdot6=900$ (cm²). Hence a side of the new photo is $\sqrt{900}=30$ (cm).

b) The old side is $\sqrt{150}$ cm. So the new side is $\dfrac{\sqrt{900}}{\sqrt{150}}=\sqrt6\approx2.45$ times as long as the old side.

Answer. a) $30$ cm; b) about $2.45$ times.

Exercise 8

The speed (m/s) of a tsunami in water $d$ metres deep is about $v=\sqrt{10d}$ .

a) Find the speed of the wave where the depth is $2.5$ m.

b) Find the speed of the wave where the depth is $4000$ m.

c) At the speed in part b, how many kilometres does the wave travel in $1$ hour?

What you need and how to start

What you need: $\sqrt{ab}=\sqrt a\cdot\sqrt b$ ; distance equals speed times time.

Steps:

Multiply under the root sign: $10\cdot2.5$ and $10\cdot4000$ .
Convert $1$ hour $=3600$ seconds.

Show solution

a) Substituting $d=2.5$ into the formula, we get $v=\sqrt{25}=5$ (m/s).

b) Substituting $d=4000$ into the formula, we get $v=\sqrt{40000}=200$ (m/s).

c) We have $1$ hour $=3600$ seconds. The distance travelled is $200\cdot3600=720000$ (m) $=720$ (km).

Answer. a) $5$ m/s; b) $200$ m/s; c) about $720$ km.

Exercise 9

Find $x\ge0$ given that $\sqrt{2x}\cdot\sqrt8=12$ .

What you need and how to start

What you need: $\sqrt{2x}\cdot\sqrt8=\sqrt{16x}=4\sqrt x$ .

Steps:

Combine the left side into $4\sqrt x$ .
Find $\sqrt x$ , then $x$ .

Show solution

We have $\sqrt{2x}\cdot\sqrt8=\sqrt{16x}=4\sqrt x$ . Hence $4\sqrt x=12\iff\sqrt x=3\iff x=9$ (which satisfies the condition).

Answer. $x=9$ .

Exercise 10 Challenge

Without a calculator, evaluate $A=\sqrt{3+\sqrt5}\cdot\sqrt{3-\sqrt5}$ .

What you need and how to start

What you need: $\sqrt a\cdot\sqrt b=\sqrt{ab}$ when $a$ , $b\ge0$ ; the identity $(m+n)(m-n)=m^2-n^2$ .

Steps:

Check that $3-\sqrt5\ge0$ .
Combine into one root and use the identity.

Show solution

Since $\sqrt5<\sqrt9=3$ , $3-\sqrt5>0$ . Hence $A=\sqrt{\left(3+\sqrt5\right)\left(3-\sqrt5\right)}=\sqrt{9-5}=\sqrt4=2$ .

Answer. $A=2$ .

a) $A=10$ .	True	False
b) $B=49$ .	True	False
c) $A\cdot B=70$ .	True	False
d) $A-B=\sqrt{51}$ .	True	False

The square root of a product

Explore: compare $\sqrt{ab}$ with $\sqrt a\cdot\sqrt b$

The square root of a quotient

Real-life applications

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Square roots with multiplication and division

The square root of a product

Explore: compare ab\sqrt{ab}ab​ with a⋅b\sqrt a\cdot\sqrt ba​⋅b​

The square root of a quotient

Real-life applications

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Explore: compare $\sqrt{ab}$ with $\sqrt a\cdot\sqrt b$