Radical Expressions and Equations

If you have made it this far in algebra, you have already encountered square roots. You know that $\sqrt{9} = 3$ because $3 \times 3 = 9$. But what happens when things get more complicated? What if you need to simplify $\sqrt{48}$, or add $\sqrt{5} + \sqrt{5}$, or solve an equation like $\sqrt{x + 3} = 5$? That is what this chapter is all about.

Here is a reassuring thought: radicals are just exponents in disguise. When you take a square root, you are asking “what number, when squared, gives me this?” In other words, you are undoing a square. And just like you learned rules for working with exponents, there are parallel rules for working with radicals. Once you see the patterns, radicals become much less mysterious.

Core Concepts

Square Roots Review

Let us start with what you already know. The square root of a number is a value that, when multiplied by itself, gives you that number.

$$\sqrt{25} = 5 \quad \text{because} \quad 5 \times 5 = 25$$

Every positive number actually has two square roots: a positive one and a negative one. For instance, both 5 and -5 square to give 25. However, when we write $\sqrt{25}$, we mean specifically the principal (positive) square root. If we want both roots, we write $\pm\sqrt{25} = \pm 5$.

Quick review of perfect squares you should know:

$n$	1	2	3	4	5	6	7	8	9	10	11	12
$n^2$	1	4	9	16	25	36	49	64	81	100	121	144

Simplifying Radicals

A radical is in simplest form when there are no perfect square factors left under the radical sign. The key insight is that you can “pull out” perfect squares from under a radical.

Why does this work? It comes from a fundamental property:

$$\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$$

This means you can break apart a square root of a product into a product of square roots. So if you spot a perfect square hiding inside your radicand, you can extract it.

For example, to simplify $\sqrt{12}$:

• Factor 12 to find perfect squares: $12 = 4 \times 3$
• Apply the property: $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3}$
• Simplify the perfect square: $= 2\sqrt{3}$

The goal is always to find the largest perfect square factor. This gives you the most simplified form in one step.

Product and Quotient Properties of Radicals

These two properties are your best friends when working with radicals.

Product Property: $$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$$

This works both ways. You can combine radicals into one, or split one radical into several.

Quotient Property: $$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \quad \text{(where } b \neq 0\text{)}$$

Again, this works in both directions. You can combine a fraction of radicals into a single radical, or split a radical of a fraction into separate radicals.

Adding and Subtracting Radicals

Here is where many students get tripped up. You cannot simply add $\sqrt{2} + \sqrt{3}$ and get $\sqrt{5}$. That is not how radicals work.

Think of it this way: $\sqrt{2}$ and $\sqrt{3}$ are like apples and oranges. You cannot combine them into one term. However, you can combine radicals that have the same radicand, just like you combine like terms in algebra.

$$3\sqrt{5} + 7\sqrt{5} = 10\sqrt{5}$$

This works exactly like $3x + 7x = 10x$. The $\sqrt{5}$ acts like a variable that you are collecting.

Important: Sometimes radicals that look different are actually like terms in disguise. You might need to simplify first:

$$\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$$

Multiplying Radicals

Multiplying radicals is straightforward thanks to the product property. You can multiply the numbers under the radicals together:

$$\sqrt{3} \cdot \sqrt{12} = \sqrt{3 \cdot 12} = \sqrt{36} = 6$$

When you have coefficients (numbers in front), multiply those separately:

$$2\sqrt{3} \cdot 5\sqrt{2} = (2 \cdot 5)(\sqrt{3} \cdot \sqrt{2}) = 10\sqrt{6}$$

Rationalizing Denominators

In mathematics, we prefer not to leave radicals in the denominator of a fraction. The process of removing them is called rationalizing the denominator.

Why do we bother? Historically, it made calculations easier (try dividing by $\sqrt{2}$ by hand). Today, it is mostly convention, but it also makes it easier to compare expressions and recognize when two answers are equivalent.

Basic technique: Multiply both the numerator and denominator by the radical in the denominator.

$$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$

This works because $\sqrt{2} \cdot \sqrt{2} = 2$, which eliminates the radical from the denominator.

Solving Radical Equations

A radical equation is an equation where the variable appears under a radical sign. To solve these, you need to isolate the radical and then “undo” it by raising both sides to the appropriate power.

For square roots, you square both sides:

$$\sqrt{x} = 5$$ $$(\sqrt{x})^2 = 5^2$$ $$x = 25$$

The process:

1. Isolate the radical on one side of the equation
2. Raise both sides to the power that matches the index (square both sides for square roots)
3. Solve the resulting equation
4. Check your answer in the original equation

That last step is crucial. Squaring both sides of an equation can introduce solutions that do not actually work in the original equation. These false solutions are called extraneous solutions.

Checking for Extraneous Solutions

When you square both sides of an equation, you might create solutions that satisfy the squared equation but not the original. This happens because squaring can make negative numbers positive.

For example, suppose squaring leads you to $x = 4$. You must substitute back:

• Original: $\sqrt{x + 5} = x - 1$
• Check: $\sqrt{4 + 5} = 4 - 1$, which gives $\sqrt{9} = 3$, so $3 = 3$ checks out.

But if squaring led you to $x = -1$:

• Check: $\sqrt{-1 + 5} = -1 - 1$, which gives $\sqrt{4} = -2$, so $2 = -2$ is false.

The second solution is extraneous; it does not work in the original equation even though it came from valid algebra.

Notation and Terminology

Term	Meaning	Example
Radical	Root symbol	$\sqrt{x}$, $\sqrt[3]{x}$
Radicand	Expression under the radical	In $\sqrt{5x}$, the radicand is $5x$
Index	The small number indicating which root	In $\sqrt[3]{8}$, the index is 3
Simplest form	No perfect square factors under radical	$\sqrt{12} = 2\sqrt{3}$
Like radicals	Radicals with the same radicand	$3\sqrt{5}$ and $7\sqrt{5}$
Rationalize	Remove radical from denominator	$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Extraneous solution	Solution from algebra that does not work in original	Must always check

Examples

Key Rules

Here is a summary of the essential rules for working with radicals:

Product Property: $$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$$

Quotient Property: $$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$

Adding/Subtracting Like Radicals: $$a\sqrt{c} + b\sqrt{c} = (a + b)\sqrt{c}$$

Simplifying Radicals: Factor out perfect squares: $\sqrt{a^2 \cdot b} = a\sqrt{b}$

Rationalizing a Denominator: $$\frac{a}{\sqrt{b}} = \frac{a\sqrt{b}}{b}$$

Solving Radical Equations:

1. Isolate the radical
2. Raise both sides to the power matching the index
3. Solve the resulting equation
4. Check all solutions in the original equation

Real-World Applications

The Distance Formula

One of the most common applications of square roots is finding distance. The distance formula, which comes from the Pythagorean theorem, calculates the straight-line distance between two points:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

For example, the distance between points (1, 2) and (4, 6) is: $$d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Period of a Pendulum

The time it takes for a pendulum to complete one full swing (its period) depends on its length:

$$T = 2\pi\sqrt{\frac{L}{g}}$$

where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity. A longer pendulum swings more slowly. Grandfather clocks use this principle: they need a pendulum about 1 meter long to tick once per second.

Speed Calculations

Police can estimate the speed of a car before braking by measuring skid marks. The formula involves a square root:

$$s = \sqrt{30 \cdot d \cdot f}$$

where $s$ is speed in mph, $d$ is the skid distance in feet, and $f$ is the friction coefficient of the road.

Engineering Formulas

Square roots appear throughout engineering. The natural frequency of a simple spring-mass system is:

$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

where $k$ is the spring stiffness and $m$ is the mass. Engineers use this to design everything from car suspensions to earthquake-resistant buildings.

Self-Test Problems

Summary

• Radicals are root symbols. The square root ($\sqrt{\ }$) asks “what number squared gives this?”
• To simplify a radical, factor out perfect squares: $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$
• Product property: $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$
• Quotient property: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$
• You can only add or subtract like radicals (same radicand): $3\sqrt{5} + 7\sqrt{5} = 10\sqrt{5}$
• To rationalize a denominator, multiply top and bottom by the radical: $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
• To solve radical equations, isolate the radical, raise both sides to the appropriate power, then solve
• Always check your answers in the original equation. Squaring can introduce extraneous solutions that do not actually work
• Radicals appear everywhere in real applications: distance formulas, pendulum motion, engineering, and physics