Simplifying Radicals

🟩 Simplifying Radicals

Part 1 of 5 — Square Roots & Perfect Squares

Topics in This Part

🔑 Key Concept: A radical like $\sqrt{n}$ asks "what number, multiplied by itself, gives $n$?" Before we can simplify radicals, we need to know our perfect squares cold — that's Part 1.

What a Square Root Means

The square root symbol $\sqrt{\phantom{n}}$ (the radical) undoes squaring. The number inside is called the radicand.

$\sqrt{n} = b \quad\text{means}\quad b^2 = n \;\;(\text{with } b \ge 0)$

So $\sqrt{25} = 5$ because $5^2 = 25$, and $\sqrt{49} = 7$ because $7^2 = 49$.

Parts of a Radical

💡 The principal (main) square root is the non-negative one. We write $\sqrt{36} = 6$, not $\pm 6$ — the $\pm$ only appears when you solve an equation like $x^2 = 36$.

Evaluate the Roots 🧮

Each radicand is a perfect square. Enter the whole-number value.

1) $\sqrt{16} = \,?$ 2) $\sqrt{121} = \,?$ 3) $\sqrt{100} = \,?$

Perfect Squares You Should Know

A perfect square is a whole number whose square root is also a whole number. Memorizing these makes simplifying radicals fast:

So $\sqrt{1}, \sqrt{4}, \sqrt{9}, \sqrt{16}, \dots, \sqrt{144}$ all come out as whole numbers.

🔑 Key Idea: If the radicand is a perfect square, the radical simplifies to a whole number. If it isn't, we leave a radical in the answer — but we can often make it smaller, which is the whole point of this lesson.

Concept Check 🎯

When Is a Radical "Already Simplified"?

A square-root radical is in simplest form when its radicand has no perfect-square factor other than $1$.

⚠️ A radical like $\sqrt{12}$ is not wrong, but it is not finished. In Part 2 we'll learn the rule that turns $\sqrt{12}$ into $2\sqrt{3}$.

Concept Check 🎯

🟩 Simplifying Radicals

Part 2 of 5 — The Product Property

🔑 The Idea: Split the radicand into a perfect-square factor times "the rest," then take the square root of the perfect square and move it out front.

The Product Property of Radicals

For any non-negative numbers $a$ and $b$:

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

This lets us break a radical apart. The trick is to break off the largest perfect-square factor.

Worked Example: $\sqrt{12}$

The largest perfect-square factor of $12$ is $4$ (since $12 = 4\cdot 3$):

$\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4}\cdot\sqrt{3} = 2\sqrt{3}$

✅ Check: $(2\sqrt{3})^2 = 2^2 \cdot 3 = 4\cdot 3 = 12$ ✓. The number that comes out front is called the coefficient.

Find the Perfect-Square Factor 🔽

For each radical, choose the largest perfect square that divides the radicand.

Putting It Together

Once you have the largest perfect-square factor, finish in one line.

Worked Example: $\sqrt{50}$

$50 = 25\cdot 2$, and $25$ is a perfect square:

$\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25}\cdot\sqrt{2} = 5\sqrt{2}$

Worked Example: $\sqrt{72}$

The largest perfect-square factor of $72$ is $36$ (since $72 = 36\cdot 2$):

$\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}$

⚠️ Pull out the LARGEST perfect square. If you only spot $\sqrt{72} = \sqrt{4\cdot 18} = 2\sqrt{18}$, you're not done — $18 = 9\cdot 2$, so $2\sqrt{18} = 2\cdot 3\sqrt{2} = 6\sqrt{2}$. Same answer, just more steps.

Concept Check 🎯

Reading the Answer

A simplified radical has two pieces:

$\underbrace{a}_{\text{coefficient}}\,\sqrt{\underbrace{b}_{\text{radicand}}}$

The coefficient $a$ is whatever escaped the radical; the radicand $b$ is what's left inside (with no perfect-square factor remaining).

💡 In the next drill, you'll report just the coefficient $a$ — the number standing in front of the radical.

Simplify These Radicals 🧮

Write each in the form $a\sqrt{b}$. Enter the coefficient $a$ (the number out front).

1) $\sqrt{8} = a\sqrt{2},\quad a = \,?$ 2) $\sqrt{48} = a\sqrt{3},\quad a = \,?$ 3) $\sqrt{200} = a\sqrt{2},\quad a = \,?$

🟩 Simplifying Radicals

Part 3 of 5 — Variables Under the Radical

🔑 Why it works: A variable raised to an even power is a perfect square. For example $x^2 = (x)^2$ and $x^6 = (x^3)^2$, so $\sqrt{x^6} = x^3$.

Throughout this part we assume variables represent non-negative numbers, so $\sqrt{x^2} = x$.

Even Powers Come Out Whole

To square-root a variable power, halve the exponent:

$\sqrt{x^{2n}} = x^{n}$

💡 Even exponent → comes out cleanly. The exponent inside is doubled; outside it is halved.

Halve the Exponent 🔽

Every exponent here is even, so each variable comes out cleanly.

Odd Powers Leave One Behind

If the exponent is odd, split off one factor so the rest is even.

Worked Example: $\sqrt{x^7}$

Write $x^7 = x^6 \cdot x$ (largest even power is $x^6$):

$\sqrt{x^7} = \sqrt{x^6 \cdot x} = \sqrt{x^6}\cdot\sqrt{x} = x^3\sqrt{x}$

Worked Example: $\sqrt{50x^3}$

Combine numbers and variables. $50 = 25\cdot 2$ and $x^3 = x^2\cdot x$:

$\sqrt{50x^3} = \sqrt{25\cdot 2 \cdot x^2 \cdot x} = \sqrt{25}\,\sqrt{x^2}\,\sqrt{2x} = 5x\sqrt{2x}$

⚠️ Everything with an even power (and every perfect square) leaves the radical; whatever is left — the odd leftover — stays inside.

Concept Check 🎯

A Three-Step Checklist

For any radical mixing numbers and variables, work through it in order:

🔑 Whatever comes out is the coefficient; whatever stays in is the radicand. The next drill checks each piece separately.

Coefficients & Exponents 🧮

Simplify each, then answer about the result.

1) $\sqrt{x^{12}} = x^{?}$ — enter the exponent. 2) $\sqrt{49x^6} = 7x^{?}$ — enter the exponent. 3) $\sqrt{8x^2} = a\,x\sqrt{2}$ — enter the coefficient $a$.

🟩 Simplifying Radicals

Part 4 of 5 — Adding, Multiplying & Rationalizing

🔑 Big Picture: Radicals behave a lot like variables. You can only add "like radicals" (same radicand), you multiply straight across, and a radical in a denominator must be cleared out.

Adding & Subtracting Radicals

You can combine radicals only when the radicand is identical — just like combining like terms ($3x + 5x = 8x$).

$3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2} \qquad 7\sqrt{5} - 2\sqrt{5} = 5\sqrt{5}$

But $\sqrt{2} + \sqrt{3}$ cannot be combined — different radicands.

The Catch: Simplify FIRST

Sometimes radicals look different but become "like" after simplifying:

$\sqrt{8} + \sqrt{18} = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$

⚠️ Never add the radicands themselves. $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$, not $\sqrt{25} = 5$.

Concept Check 🎯

Multiplying Radicals

Multiply coefficients with coefficients and radicands with radicands, then simplify:

$a\sqrt{m}\cdot b\sqrt{n} = (a\cdot b)\sqrt{m\cdot n}$

Worked Example: $\sqrt{6}\cdot\sqrt{2}$

$\sqrt{6}\cdot\sqrt{2} = \sqrt{12} = 2\sqrt{3}$

Worked Example: $2\sqrt{3}\cdot 5\sqrt{6}$

$2\sqrt{3}\cdot 5\sqrt{6} = (2\cdot 5)\sqrt{3\cdot 6} = 10\sqrt{18} = 10\cdot 3\sqrt{2} = 30\sqrt{2}$

💡 A useful special case: $\sqrt{a}\cdot\sqrt{a} = a$. For instance $\sqrt{7}\cdot\sqrt{7} = 7$.

Multiply & Simplify 🧮

Write each product in the form $a\sqrt{b}$ (fully simplified).

1) $\sqrt{3}\cdot\sqrt{12} = \,?$ (a whole number — enter it) 2) $\sqrt{2}\cdot\sqrt{10} = a\sqrt{5}$, enter $a$. 3) $\sqrt{5}\cdot\sqrt{5} = \,?$ (a whole number)

Rationalizing the Denominator

A radical is not in simplest form if it sits in a denominator. To fix this, multiply the top and bottom by that radical — this clears the root downstairs because $\sqrt{a}\cdot\sqrt{a} = a$.

Worked Example: $\dfrac{1}{\sqrt{2}}$

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

Worked Example: $\dfrac{6}{\sqrt{3}}$

$\frac{6}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

🔑 Goal: No radical in any denominator and no perfect-square factor in any radicand. That is the official "simplest form" for a radical expression.

Rationalize & Combine 🔽

🟩 Simplifying Radicals

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) evaluate perfect-square roots, (2) simplify numeric radicals with the product property, (3) handle variables under the radical, and (4) add, multiply, and rationalize. Let's put it all together.

Quick Reference

⚠️ Simplest form means: no perfect-square factor left inside, and no radical in a denominator.

Mixed Practice 🎯

One Problem, Every Skill

A single expression can touch all four skills at once. Take $\sqrt{32x^4}$:

$\sqrt{32x^4} = \sqrt{16\cdot 2}\,\sqrt{x^4} = 4x^2\sqrt{2}$

Pull out the perfect-square number ($16$), halve the even exponent ($x^4 \to x^2$), and leave the rest ($2$) inside. The next drill mixes simplifying, multiplying, and rationalizing in one set.

Mixed Drill 🧮

1) $\sqrt{32x^4} = a\,x^2\sqrt{2}$ — enter the coefficient $a$. 2) $\sqrt{6}\cdot\sqrt{6} = \,?$ (whole number) 3) $\dfrac{10}{\sqrt{5}} = a\sqrt{5}$ — enter $a$.

Before the Exit Quiz

Keep the two rules of simplest form in mind as you finish:

✅ No perfect-square factor left inside any radical. ✅ No radical left in any denominator.

If both are true, your answer is fully simplified. Three questions to go.

Exit Quiz ✅

Answer all three to finish the lesson.