Square roots and square-root expressions

Mr Hùng wants to tile a square drying yard with an area of $49$ m². How long is each side? What if the yard has an area of $50$ m²?

The first question is easy: $7\cdot7=49$ , so each side is $7$ m. The second needs a number whose square is $50$ . This lesson helps us name, write and calculate numbers like that.

Square roots of a non-negative number

Explore: which numbers square to $16$ ?

We square a few numbers:

$x$	$-5$	$-4$	$-3$	$0$	$3$	$4$	$5$
$x^2$	$25$	$16$	$9$	$0$	$9$	$16$	$25$

Two numbers square to $16$ : $4$ and $-4$ . Only one number squares to $0$ , namely $0$ . No number squares to $-9$ , because the square of every number is non-negative.

The picture shows why a positive number has two square roots: the line $y=5$ meets the graph of $y=x^2$ at two points whose $x$ -coordinates are opposites.

The two square roots of

5

For a number like $50$ , no rational number squares to $50$ . Then we use a calculator to get an approximate value.

To compare two principal square roots we use this property: for non-negative $a$ and $b$ , $a<b \iff \sqrt{a}<\sqrt{b}$ . For example, $\sqrt{50}<\sqrt{64}=8$ .

The identity $\sqrt{A^2}=|A|$

We have $\sqrt{3^2}=\sqrt9=3$ . But $\sqrt{(-3)^2}=\sqrt9=3$ , not $-3$ . The result is always non-negative: it is exactly the absolute value.

Why? The number $|a|$ is non-negative and $|a|^2=a^2$ . So $|a|$ is the principal square root of $a^2$ .

Square-root expressions

In real life we often meet the square root of an expression with letters. For example, a pebble dropped from a height of $h$ metres hits the ground after about $t=\sqrt{\dfrac{h}{5}}$ seconds.

The height is more than twice as great, but the time grows by only $1$ second.

Finding when an expression is defined means solving a linear inequality in one unknown, as in the lesson linear inequalities in one unknown.

Example When is it defined?

Find the condition on $x$ for each expression to be defined: a) $\sqrt{2x-6}$ ; b) $\sqrt{5-x}$ ; c) $\sqrt{x^2+1}$ .

Solution. Each expression is defined when the expression under the root sign is non-negative (in b), multiplying both sides by $-1$ reverses the inequality sign). We have

$\begin{aligned} \text{a) }& 2x-6\ge0\iff 2x\ge6\iff x\ge3;\\ \text{b) }& 5-x\ge0\iff -x\ge-5\iff x\le5;\\ \text{c) }& x^2+1\ge1>0 \text{ for every } x. \end{aligned}$

So a) $x\ge3$ ; b) $x\le5$ ; c) the expression is defined for every real number $x$ .

When the expression under the root sign is a square, we remove the root using the absolute value, then use the given condition to remove the absolute value.

Watch out Common mistakes

Writing $\sqrt{16}=\pm4$ . Why it is wrong: the symbol $\sqrt{16}$ means the principal square root, which is never negative. Right: $\sqrt{16}=4$ ; the square roots of $16$ are $4$ and $-4$ .
Writing $\sqrt{a^2}=a$ for every $a$ . Wrong: $\sqrt{(-3)^2}=-3$ . Why it is wrong: the principal square root is non-negative. Right: $\sqrt{(-3)^2}=|-3|=3$ .
Splitting the root of a sum. Wrong: $\sqrt{9+16}=\sqrt9+\sqrt{16}=7$ . Why it is wrong: a square root does not split over addition. Right: $\sqrt{9+16}=\sqrt{25}=5$ .
Forgetting to reverse the inequality. Wrong: $\sqrt{4-2x}$ is defined when $x\ge2$ . Why it is wrong: dividing both sides of $-2x\ge-4$ by a negative number reverses the sign. Right: $x\le2$ .

Key terms

English	Tiếng Việt
square root	căn bậc hai
principal square root	căn bậc hai số học
square-root expression	căn thức bậc hai
expression under the root sign	biểu thức lấy căn
condition for the expression to be defined	điều kiện xác định
absolute value	giá trị tuyệt đối
irrational number	số vô tỉ
calculator	máy tính cầm tay

Summary

$x$ is a square root of $a\ge0$ when $x^2=a$ . A positive $a$ has two square roots $\pm\sqrt{a}$ ; $0$ has one; a negative number has none.
$\sqrt{a}$ is the principal square root: $\sqrt{a}\ge0$ and $\left(\sqrt{a}\right)^2=a$ .
For $a$ , $b\ge0$ : $a<b\iff\sqrt{a}<\sqrt{b}$ .
$\sqrt{A^2}=|A|$ : it equals $A$ when $A\ge0$ and $-A$ when $A<0$ .
$\sqrt{A}$ is defined $\iff A\ge0$ .

Practice

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

Exercise 1

Find the square roots of: a) $121$ ; b) $0.49$ ; c) $\dfrac{25}{36}$ .

What you need and how to start

What you need: a positive number $a$ has two square roots, $\sqrt{a}$ and $-\sqrt{a}$ .

Steps:

Find the positive number whose square is the given number.
Add its opposite.

Show solution

We have $11^2=121$ , $0.7^2=0.49$ , $\left(\dfrac56\right)^2=\dfrac{25}{36}$ .

Answer. a) $\pm11$ ; b) $\pm0.7$ ; c) $\pm\dfrac56$ .

Exercise 2

Evaluate: a) $\sqrt{0.64}+\sqrt{1.21}$ ; b) $\left(\sqrt7\right)^2-\sqrt{(-6)^2}$ .

What you need and how to start

What you need: $\left(\sqrt{a}\right)^2=a$ for $a\ge0$ ; $\sqrt{a^2}=|a|$ .

Steps:

Take each square root.
Add or subtract.

Show solution

a) We have $\sqrt{0.64}+\sqrt{1.21}=0.8+1.1=1.9$ .

b) We have $\left(\sqrt7\right)^2-\sqrt{(-6)^2}=7-|-6|=7-6=1$ .

Answer. a) $1.9$ ; b) $1$ .

Exercise 3

Compare: a) $6$ and $\sqrt{35}$ ; b) $\sqrt{10}-1$ and $2$ .

What you need and how to start

What you need: for $a$ , $b\ge0$ : $a<b\iff\sqrt{a}<\sqrt{b}$ .

Steps:

Write the whole number as a square root: $6=\sqrt{36}$ , $3=\sqrt9$ .
Compare the numbers under the root signs.

Show solution

a) We have $6=\sqrt{36}$ . Since $36>35$ , $\sqrt{36}>\sqrt{35}$ , hence $6>\sqrt{35}$ .

b) We have $3=\sqrt9$ . Since $10>9$ , $\sqrt{10}>3$ . So $\sqrt{10}-1>2$ .

Answer. a) $6>\sqrt{35}$ ; b) $\sqrt{10}-1>2$ .

Exercise 4

Find the condition on $x$ for each expression to be defined: a) $\sqrt{3x-12}$ ; b) $\sqrt{7-2x}$ ; c) $\sqrt{x^2+4}$ .

What you need and how to start

What you need: $\sqrt{A}$ is defined when $A\ge0$ .

Steps:

Write the inequality $A\ge0$ .
Solve it; remember to reverse the sign when dividing by a negative number.

Show solution

a) $\sqrt{3x-12}$ is defined when $3x-12\ge0\iff 3x\ge12\iff x\ge4$ .

b) $\sqrt{7-2x}$ is defined when $7-2x\ge0\iff -2x\ge-7\iff x\le3.5$ .

c) Since $x^2+4\ge4>0$ for every $x$ , $\sqrt{x^2+4}$ is defined for every $x$ .

Answer. a) $x\ge4$ ; b) $x\le3.5$ ; c) every $x$ .

Exercise 5

Simplify the following expressions: a) $\sqrt{\left(\sqrt3-1\right)^2}$ ; b) $\sqrt{\left(3-\sqrt{11}\right)^2}$ .

What you need and how to start

What you need: $\sqrt{A^2}=|A|$ .

Steps:

Find the sign of the number inside (compare $\sqrt3$ with $1$ , and $\sqrt{11}$ with $3$ ).
Remove the absolute value.

Show solution

a) Since $\sqrt3>\sqrt1=1$ , $\sqrt3-1>0$ . Hence $\sqrt{\left(\sqrt3-1\right)^2}=\left|\sqrt3-1\right|=\sqrt3-1$ .

b) Since $\sqrt{11}>\sqrt9=3$ , $3-\sqrt{11}<0$ . Hence $\sqrt{\left(3-\sqrt{11}\right)^2}=\left|3-\sqrt{11}\right|=\sqrt{11}-3$ .

Answer. a) $\sqrt3-1$ ; b) $\sqrt{11}-3$ .

Exercise 6

A square rug has an area of $30$ m². Find its perimeter (rounded to one decimal place).

What you need and how to start

What you need: a square of area $S$ has side $\sqrt{S}$ .

Steps:

The side is $\sqrt{30}$ m.
The perimeter is $4$ times the side; use the calculator and round only at the end.

Show solution

The side of the rug is $\sqrt{30}$ m. Hence its perimeter is $4\sqrt{30}\approx21.9$ (m).

Answer. About $21.9$ m.

Exercise 7

Use the formula $t=\sqrt{\dfrac{h}{5}}$ (seconds) for an object falling freely from a height of $h$ metres.

a) A ball is dropped from a height of $80$ m. How long does it take to hit the ground?

b) A marble takes exactly $2.5$ seconds to fall. From what height was it dropped?

What you need and how to start

What you need: $t=\sqrt{\dfrac{h}{5}}\iff t^2=\dfrac{h}{5}\iff h=5t^2$ (for $t\ge0$ ).

Steps:

Part a: substitute $h=80$ .
Part b: use $h=5t^2$ .

Show solution

a) Substituting $h=80$ into the formula, we get $t=\sqrt{\dfrac{80}{5}}=\sqrt{16}=4$ .

b) From $t=\sqrt{\dfrac h5}$ we get $h=5t^2=5\cdot2.5^2=31.25$ .

Answer. a) $4$ seconds; b) $31.25$ m.

Exercise 8

Simplify the expression $Q=\sqrt{x^2-6x+9}-x$ for $x\ge3$ .

What you need and how to start

What you need: the identity $(a-b)^2=a^2-2ab+b^2$ (grade 8) and $\sqrt{A^2}=|A|$ .

Steps:

Write $x^2-6x+9=(x-3)^2$ .
For $x\ge3$ , $|x-3|=x-3$ .

Show solution

We have $Q=\sqrt{(x-3)^2}-x=|x-3|-x$ . Since $x\ge3$ , $|x-3|=x-3$ . Hence $Q=x-3-x=-3$ .

Answer. $Q=-3$ .

Exercise 9

Consider the expression $A=\sqrt{4x-8}$ .

a) Find the condition for $A$ to be defined.

b) Find the value of $A$ when $x=6$ .

c) Find $x$ such that $A=6$ .

What you need and how to start

What you need: $\sqrt{B}$ is defined when $B\ge0$ ; for $c\ge0$ , $\sqrt{B}=c\iff B=c^2$ .

Steps:

Solve $4x-8\ge0$ .
Substitute $x=6$ .
Solve $4x-8=36$ and check the condition.

Show solution

a) $A$ is defined when $4x-8\ge0\iff x\ge2$ .

b) Substituting $x=6$ (which satisfies the condition) into $A$ , we get $A=\sqrt{24-8}=\sqrt{16}=4$ .

c) We have $A=6\iff 4x-8=36\iff x=11$ (which satisfies the condition).

Answer. a) $x\ge2$ ; b) $A=4$ ; c) $x=11$ .

Exercise 10 Challenge

Find $x$ given that $\sqrt{x^2}=2x-3$ .

What you need and how to start

What you need: $\sqrt{x^2}=|x|\ge0$ , so the right side must also be non-negative.

Steps:

Condition: $2x-3\ge0$ , that is, $x\ge1.5$ .
Then $x>0$ , so $|x|=x$ ; solve $x=2x-3$ .

Show solution

Since $\sqrt{x^2}=|x|\ge0$ , we need $2x-3\ge0$ , that is, $x\ge1.5$ . Then $x>0$ , so $|x|=x$ . Hence $x=2x-3\iff x=3$ (which satisfies the condition).

Answer. $x=3$ .

a) $A$ is defined when $x\ge-5$ .	True	False
b) When $x=3$ , $A=4$ .	True	False
c) When $x=-3$ , $A=-2$ .	True	False
d) $A=0$ when $x=5$ .	True	False

Square roots of a non-negative number

Explore: which numbers square to $16$ ?

The identity $\sqrt{A^2}=|A|$

Square-root expressions

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Square roots and square-root expressions

Square roots of a non-negative number

Explore: which numbers square to 161616?

The identity A2=∣A∣\sqrt{A^2}=|A|A2​=∣A∣

Square-root expressions

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

Explore: which numbers square to $16$ ?

The identity $\sqrt{A^2}=|A|$