International Baccalaureate · MYP 4

Radicals and Surds

⌚ ~154 min ▢ 17 blocks ✓ 84 exercises ➣ Prerequisites : Algebraic Expansion, Square and Cube Roots, Pythagoras' Theorem

I Calculating nth Roots

Definition — Cube Root

The cube root of a number $a$, written $\sqrt[3]{a}$, is the number which, when cubed, gives $a$: $$\left(\sqrt[3]{a}\right)^3 = a$$

Note

Unlike square roots, cube roots are defined for all real numbers, including negatives. For example, $\sqrt[3]{-27} = -3$ because $(-3) \times (-3) \times (-3) = -27$.

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Definition — Nth Root

The concept of a root can be generalised. For a positive integer $n$, the nth root of a real number $a$, written $\sqrt[n]{a}$, is a number which, when raised to the power of $n$, gives $a$, chosen as follows: for an odd $n$, it is the only real number whose $n$th power is $a$; for an even $n$, it exists only when $a \ge 0$, and it is the non-negative number whose $n$th power is $a$. Whenever it exists, $$\left(\sqrt[n]{a}\right)^n = a.$$ The rules for their domains depend on whether the index $n$ is even or odd.

Even roots (e.g.\ $\sqrt[2]{a}, \sqrt[4]{a}, \dots$): an even root of a negative number is not a real number. For even $n$, $\sqrt[n]{a}$ is only defined for $\boldsymbol{a \ge 0}$, and it denotes the non-negative number whose $n$th power is $a$ (e.g.\ $\sqrt[4]{16} = 2$, not $-2$).
Odd roots (e.g.\ $\sqrt[3]{a}, \sqrt[5]{a}, \dots$): an odd root is defined for all real numbers $a$.

Example

Calculate each root, or explain why it is not a real number:

$\sqrt[4]{625}$
$\sqrt[5]{-243}$
$\sqrt[4]{-625}$

Answer

$5^4 = 5 \times 5 \times 5 \times 5 = 625$ and $5 \geq 0$, so $\sqrt[4]{625} = 5$. The number $-5$ also has $625$ as its fourth power, but the fourth root is the non-negative one.
The index $5$ is odd, so the fifth root of the negative number $-243$ exists. Since $(-3)^5 = (-3) \times (-3) \times (-3) \times (-3) \times (-3) = -243$, we have $\sqrt[5]{-243} = -3$.
The index $4$ is even and $-625 < 0$. An even power of a real number is never negative, so no real number has $-625$ as its fourth power: $\sqrt[4]{-625}$ is not a real number.

Skills to practice

Understanding nth Roots
Calculating nth Roots

II Using the Laws of Radicals

Definition — Square root

The square root of a non-negative number $a$ (that is, $a \ge 0$), written as $\sqrt{a}$, is the non-negative number that, when multiplied by itself, gives $a$. $$ \left(\sqrt{a}\right)^2 = a $$

Definition — Perfect Squares

A perfect square is an integer that is the square of an integer. The square root of a perfect square is an integer.

Expressions involving square roots are called radical expressions (or simply radicals). To simplify and manipulate these expressions, we use a set of important laws.

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Proposition — Multiplication Law

For any real numbers $a,b \ge 0$: $$\textcolor{colorprop} {\sqrt{a} \times \sqrt{b} = \sqrt{ab}} $$

Proof

By definition, squaring a square root gives the original number. Look at the square of the expression $\sqrt{a} \times \sqrt{b}$: $$\begin{aligned}[t] \left(\sqrt{a} \times \sqrt{b}\right)^2 &= (\sqrt{a} \times \sqrt{b}) \times (\sqrt{a} \times \sqrt{b}) && \text{(definition of squaring)} \\ &= (\sqrt{a} \times \sqrt{a}) \times (\sqrt{b} \times \sqrt{b}) && \text{(commutativity)} \\ &= a \times b && \text{($(\sqrt{a})^2 = a$ and $(\sqrt{b})^2 = b$)} \\ &= ab. \end{aligned}$$ Moreover $\sqrt{a} \times \sqrt{b} \ge 0$, as a product of two non-negative numbers. So $\sqrt{a} \times \sqrt{b}$ is the non-negative number whose square is $ab$: it is the square root of $ab$.

Example

Check that $\sqrt{4} \times \sqrt{9} = \sqrt{36}$ by calculating each side.

Answer

Each side is calculated on its own, without using the Multiplication Law. $$\begin{aligned}[t] \sqrt{4} \times \sqrt{9} &= 2 \times 3 && \text{($2^2 = 4$ and $3^2 = 9$)} \\ &= 6 \\ \sqrt{36} &= 6. && \text{($6^2 = 36$)} \end{aligned}$$ Both sides are equal to $6$: the equality is true, as the Multiplication Law states, since $4 \times 9 = 36$.

Proposition — Square Root of a Square

For any real number $a \ge 0$: $$ \textcolor{colorprop}{\sqrt{a^2} = a} $$

Proof

By definition, $\sqrt{a^2}$ is the non-negative number whose square is $a^2$. The number $a$ is non-negative and its square is $a^2$. So: $$\sqrt{a^2} = a$$

Example

Find $\sqrt{13^2}$ and $\sqrt{0.7^2}$ without calculating the squares.

Answer

$$\begin{aligned}[t] \sqrt{13^2} &= 13 && \text{(Square Root of a Square, $13 \geq 0$)} \\ \sqrt{0.7^2} &= 0.7. && \text{(Square Root of a Square, $0.7 \geq 0$)} \end{aligned}$$ There is no need to calculate $13^2 = 169$ or $0.7^2 = 0.49$: the square root undoes the square of a non-negative number.

Proposition — Simplifying Law

For any real numbers $a,b \ge 0$: $$ \textcolor{colorprop}{\sqrt{a^2 b} = a\sqrt{b}} $$

Proof

$$\begin{aligned}[t] \sqrt{a^2 b} &= \sqrt{a^2} \times \sqrt{b} && \text{(by Multiplication Law)} \\ &= a \times \sqrt{b} && \text{(by Square Root of a Square Law)} \\ &= a\sqrt{b} \end{aligned}$$

Definition — Simplest Radical Form

A radical is written in simplest form when the number under the square root sign has no perfect-square factor other than $1$. For example, $\sqrt{12}$ is not in simplest form, because $12 = 4 \times 3$ and $4$ is a perfect square; its simplest form is $2\sqrt{3}$.

Method — Writing a Radical in Simplest Form

Let $n \geq 1$ be a whole number. To write $\sqrt{n}$ in simplest form:

Find the largest perfect square that divides $n$, and write $n = a^2 \times b$ with $a > 0$.
Split the root with the Multiplication Law: $\sqrt{n} = \sqrt{a^2} \times \sqrt{b}$.
Extract the square: $\sqrt{a^2} = a$ (since $a > 0$), so $\sqrt{n} = a\sqrt{b}$.

Example

Simplify $\sqrt{72}$.

Answer

First, find the largest perfect square factor of $72$, which is $36$. $$\begin{aligned}[t] \sqrt{72} &= \sqrt{36 \times 2} && \text{($36$ is the largest square factor)} \\ &= \sqrt{6^2 \times 2} && \text{($36 = 6^2$)} \\ &= 6\sqrt{2}. && \text{(by Simplifying Law)} \end{aligned}$$ With the smaller square factor $4$, the result $\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}$ is not yet in simplest form, because $18 = 9 \times 2$ still contains a perfect square: taking the largest square factor finishes in one step.

Proposition — Division Law

For any real numbers $a \ge 0$ and $b>0$: $$ \textcolor{colorprop}{\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}} $$

Proof

Look at the square of the expression $\dfrac{\sqrt{a}}{\sqrt{b}}$: $$\begin{aligned}[t] \left(\frac{\sqrt{a}}{\sqrt{b}}\right)^2 &= \frac{\sqrt{a}}{\sqrt{b}} \times \frac{\sqrt{a}}{\sqrt{b}} && \text{(definition of squaring)} \\ &= \frac{(\sqrt{a})^2}{(\sqrt{b})^2} && \text{(multiply the fractions)} \\ &= \frac{a}{b}. && \text{($(\sqrt{a})^2 = a$ and $(\sqrt{b})^2 = b$)} \end{aligned}$$ Moreover $\dfrac{\sqrt{a}}{\sqrt{b}} \ge 0$, as a quotient of non-negative numbers (with $\sqrt{b} > 0$). So it is the non-negative number whose square is $\dfrac{a}{b}$: it is the square root of $\dfrac{a}{b}$.

Example

Simplify $\sqrt{\dfrac{5}{9}}$.

Answer

$$\begin{aligned}[t] \sqrt{\frac{5}{9}} &= \frac{\sqrt{5}}{\sqrt{9}} && \text{(by Division Law)} \\ &= \frac{\sqrt{5}}{3}. && \text{($\sqrt{9} = 3$)} \end{aligned}$$ The numerator $5$ is not a perfect square: the Division Law still applies, and only the denominator becomes a whole number.

Surds and Exact Values

The square roots $\sqrt{2}$, $\sqrt{3}$ or $\sqrt{12} = 2\sqrt{3}$ are irrational numbers: none of them can be written as a fraction of two integers, so none of them is equal to a terminating decimal. A root of this kind, whose value is irrational, is also called a surd; the word is used for other roots too, such as $\sqrt[3]{5}$.

A number written with its roots, such as $\sqrt{12}$ or $2\sqrt{3}$, is written in exact form: both are exact values, while a calculator only gives an approximation, $\sqrt{12} \approx 3.464$.
Unless a question asks for a decimal value, an answer is given in exact form, with its roots in simplest form: $2\sqrt{3}$ rather than $\sqrt{12}$.

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Skills to practice

Writing as a Single Root
Simplifying the Square Root of a Perfect Square
Simplifying Square Roots
Simplifying Quotients of Square Roots

III Finding the Root of a Square and the Root of a Sum

Proposition — Square Root of a Square: the General Case

For every real number $a$: $$\sqrt{a^2} = \begin{cases} a & \text{if } a \geq 0,\\ -a & \text{if } a < 0. \end{cases}$$ This single result — the non-negative one of $a$ and $-a$ — is written $\lvert a \rvert$, the absolute value of $a$. In short: $$\sqrt{a^2} = \lvert a \rvert.$$ In particular, $\sqrt{x^2} = x$ is false when $x$ is strictly negative.

Proof

By definition, $\sqrt{a^2}$ is the non-negative number whose square is $a^2$. We distinguish two cases.

If $a \geq 0$: the number $a$ is non-negative and its square is $a^2$, so $\sqrt{a^2} = a$.
If $a < 0$: the number $-a$ is positive, and $(-a)^2 = a^2$, so $\sqrt{a^2} = -a$.

In both cases, $\sqrt{a^2}$ is the non-negative one of $a$ and $-a$, that is $\lvert a \rvert$.

Example

Compute $\sqrt{(-5)^2}$, and explain why the result is not $-5$.

Answer

$$\begin{aligned}[t] \sqrt{(-5)^2} &= \sqrt{25} \\ &= 5 \\ &= \lvert -5 \rvert. \end{aligned}$$ The square root of a number is non-negative by definition: it cannot equal $-5$. The rule $\sqrt{a^2} = \lvert a \rvert$ gives the correct sign automatically.

Proposition — Comparing Positive Numbers by their Squares

For all real numbers $a \geq 0$ and $b \geq 0$: $$a < b \quad \text{if and only if} \quad a^2 < b^2.$$ Two non-negative numbers are ordered like their squares: to compare them, it is enough to compare their squares.

Proof

The key is the factorisation $b^2 - a^2 = (b - a)(b + a)$.

If $a < b$: then $b - a > 0$, and $b + a > 0$ since $b > a \geq 0$. The product $(b - a)(b + a)$ is positive, so $b^2 - a^2 > 0$, that is $a^2 < b^2$.
If $a^2 < b^2$: then $b^2 > 0$, so $b > 0$ and $b + a > 0$. The product $(b - a)(b + a) = b^2 - a^2$ is positive and its second factor is positive, so $b - a > 0$, that is $a < b$.

Example

Compare $\sqrt{7}$ and $2.7$ without a calculator.

Answer

Both numbers are non-negative, so we compare their squares: $$\begin{aligned}[t] \left(\sqrt{7}\right)^2 &= 7 \\ 2.7^2 &= 7.29. \end{aligned}$$ Since $7 < 7.29$: $\quad \sqrt{7} < 2.7$.

Proposition — The Root of a Sum

For all real numbers $a > 0$ and $b > 0$: $$\sqrt{a + b} < \sqrt{a} + \sqrt{b}.$$ The square root is not additive: $\sqrt{a+b}$ is never equal to $\sqrt{a} + \sqrt{b}$ when $a$ and $b$ are positive.

Proof

The two numbers $\sqrt{a+b}$ and $\sqrt{a} + \sqrt{b}$ are positive; to compare them, we compare their squares. $$\begin{aligned}[t] \left(\sqrt{a+b}\right)^2 &= a + b \\ \left(\sqrt{a} + \sqrt{b}\right)^2 &= a + 2\sqrt{a}\sqrt{b} + b && \text{(expanding the square)}\\ &= (a + b) + 2\sqrt{ab}. && \text{(multiplication law)} \end{aligned}$$ Since $a > 0$ and $b > 0$, we have $\sqrt{ab} > 0$, so $$\left(\sqrt{a} + \sqrt{b}\right)^2 = (a+b) + 2\sqrt{ab} > \left(\sqrt{a+b}\right)^2.$$ Two positive numbers compare as their squares do, hence $\sqrt{a} + \sqrt{b} > \sqrt{a+b}$.

Example

Check the proposition with $a = 16$ and $b = 9$.

Answer

$$\begin{aligned}[t] \sqrt{16 + 9} &= \sqrt{25} = 5 \\ \sqrt{16} + \sqrt{9} &= 4 + 3 = 7 \end{aligned}$$ Indeed $5 < 7$: taking the root of the sum gives a smaller number than summing the roots.

Skills to practice

Deciding Whether a Root Equality Holds
Computing Roots of Squares

IV Adding, Subtracting and Expanding Radicals

A sum or a difference of radicals can be combined into a single term only when they are like radicals, which means they have the same index and exactly the same number under the root sign.
As in algebra, $2x+4x$ simplifies to $6x$, but $2x+4y$ cannot be simplified. In the same way, $2\sqrt{3} + 4\sqrt{3}$ can be simplified, but $2\sqrt{3} + 4\sqrt{5}$ cannot. Radicals that do not look alike may become like radicals after simplification: $\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$.
When multiplying expressions with radicals, we use the same rules as in algebra, such as the distributive law for expanding brackets, together with the laws of radicals.

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Method — Algebraic Operations

We can perform operations with radicals (square roots) in a similar way to ordinary numbers, provided we respect the radical laws. In particular:

We can add and subtract like radicals (i.e.\ the same number under the root) in the same way that we add and subtract like algebraic terms: $$ \textcolor{colorprop}{c\sqrt{a} + d\sqrt{a} = (c + d)\sqrt{a}} $$
We can use the usual rules for expanding brackets (such as the distributive law).

Example

Simplify: $6\sqrt{5} - 2\sqrt{5} + \sqrt{5}$

Answer

Just like $6x - 2x + x = 5x$, we combine the coefficients of the like radical; the last term is $\sqrt{5} = 1 \times \sqrt{5}$, with the coefficient $1$: $$\begin{aligned}[t] 6\sqrt{5} - 2\sqrt{5} + \sqrt{5} &= (6 - 2 + 1)\sqrt{5} && \text{(factor out $\sqrt{5}$)} \\ &= 5\sqrt{5}. && \text{($6 - 2 + 1 = 5$)} \end{aligned}$$

Example

Expand and simplify $\sqrt{3}(5 - \sqrt{3})$.

Answer

$$\begin{aligned}[t] \sqrt{3}(5 - \sqrt{3}) &= (\sqrt{3} \times 5) - (\sqrt{3} \times \sqrt{3}) && \text{(distributive law)} \\ &= 5\sqrt{3} - 3. && \text{($\sqrt{3} \times \sqrt{3} = 3$)} \end{aligned}$$

Example

Expand and simplify $(2 + \sqrt{3})(4 - \sqrt{3})$.

Answer

Each term of the first bracket multiplies each term of the second one, then the like radicals are collected. $$\begin{aligned}[t] (2 + \sqrt{3})(4 - \sqrt{3}) &= 2 \times 4 - 2\sqrt{3} + 4\sqrt{3} - \sqrt{3} \times \sqrt{3} && \text{(expanding two brackets)} \\ &= 8 - 2\sqrt{3} + 4\sqrt{3} - 3 && \text{($\sqrt{3} \times \sqrt{3} = 3$)} \\ &= 5 + 2\sqrt{3}. \end{aligned}$$

Example

Expand and simplify:

$(3 - \sqrt{2})^2$
$(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})$

Answer

The identities of algebra hold for radicals: $(a - b)^2 = a^2 - 2ab + b^2$ and $(a + b)(a - b) = a^2 - b^2$.

With $a = 3$ and $b = \sqrt{2}$: $$\begin{aligned}[t] (3 - \sqrt{2})^2 &= 3^2 - 2 \times 3 \times \sqrt{2} + (\sqrt{2})^2 && \text{(square of a difference)} \\ &= 9 - 6\sqrt{2} + 2 && \text{($(\sqrt{2})^2 = 2$)} \\ &= 11 - 6\sqrt{2}. \end{aligned}$$
With $a = \sqrt{5}$ and $b = \sqrt{2}$: $$\begin{aligned}[t] (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) &= (\sqrt{5})^2 - (\sqrt{2})^2 && \text{(difference of two squares)} \\ &= 5 - 2 \\ &= 3. \end{aligned}$$

The product of two numbers written with radicals can be a whole number: the radicals disappear in the difference of two squares.

Skills to practice

Adding and Subtracting Like Radicals: Level 1
Adding and Subtracting Like Radicals: Level 2
Expanding Products with Radicals

V Dividing by a Square Root

In mathematics, it is standard practice to write expressions without radicals in the denominator. The process of removing a radical from the denominator is called rationalising the denominator. This does not change the value of the expression but converts it to a conventional form which is often easier to work with.

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Method — Rationalising a Monomial Denominator

To rationalise a denominator of the form $\sqrt{b}$ (where $b>0$), multiply the numerator and the denominator by $\sqrt{b}$.$$ \frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b} $$ This works because multiplying by $\frac{\sqrt{b}}{\sqrt{b}}$ is equivalent to multiplying by 1, which does not change the value.

Example

Rationalise the denominator of $\frac{6}{\sqrt{2}}$.

Answer

We multiply the numerator and denominator by $\sqrt{2}$.$$\begin{aligned}[t]\frac{6}{\sqrt{2}} &= \frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} && \text{(multiply by 1)} \\ &= \frac{6\sqrt{2}}{2} && \text{($\sqrt{2} \times \sqrt{2} = 2$)} \\ &= \boldsymbol{3\sqrt{2}}\end{aligned}$$

Skills to practice

Writing With an Integer Denominator: Level 1
Writing With an Integer Denominator: Level 2

VI Going further

Skills to practice

Exact Lengths in a Square
Approximating Roots
Comparing and Proving with Radicals
Using Exact Values in Context
Quiz