Adding, subtracting, multiplying and dividing radical expressions. Includes combining radicals with the same radicand using the distributive property, simplifying first so that unlike radicands may turn out to match, multiplying with the product and distributive properties, the sum and difference pattern that removes a radical entirely, and rationalising a denominator that is a single radical or a sum.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 12 — Radicals and More Connections to Geometry
Operations with Radical Expressions
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-703 — the lesson these objectives are drawn from
Warm-up
Lesson 9.3 simplified single radicals. This lesson combines several of them, using rules you already know for polynomials.
Discussion prompt
Simplify three x plus x. Then try the same thing with the root of two in place of x.
Hint: Count the copies.
Answer:
\[ 3x + x = 4x, \qquad 3\sqrt{2} + \sqrt{2} = 4\sqrt{2} \]
The root of two behaves exactly like a variable: three of them plus one of them is four of them. Every rule in this lesson comes from treating a radical as a single object rather than as something to be evaluated.
Concept
Radical expressions with the same radicand can be combined using the distributive property, in the same way that like terms are combined. Only the coefficients change.
simplest form of a radical expression — A radical expression with no perfect square factors other than one in any radicand, no fractions under a radical, and no radicals in any denominator.
Radicands must match before anything can be combined.
Figure (svg): Radicals with the same radicand combined like terms
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698
Section
Section 1
Concept
The distributive property lets radicals with the same radicand be combined. Radicals with different radicands stay as separate terms.
\[ a\sqrt{c} + b\sqrt{c} = (a + b)\sqrt{c} \]
The radicand itself is never added.
Figure (svg): Radicals with the same radicand combined like terms
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698 — the distributive property applied to sums and differences of radicals, and Example 1
Picture it
Only matching ones combine.
Figure (svg): Two columns separating radicals that combine from those that do not
The rule is Chapter 2's rule about like terms with a radical in place of a variable. Nothing about radicals makes them behave differently.
Worked example
This is Example 1, part a, from the textbook.
\[ \text{Simplify } 2\sqrt{2} + \sqrt{5} - 6\sqrt{2}. \]
Group the matching radicands
Why: Two of them have root two.
\[ (2\sqrt{2} - 6\sqrt{2}) + \sqrt{5} \]
Combine the coefficients
Why: Two minus six.
\[ -4\sqrt{2} \]
Leave the odd one
Why: Root five has no partner.
\[ + \sqrt{5} \]
Write the answer
Why: Two unlike terms.
\[ -4\sqrt{2} + \sqrt{5} \]
Figure (svg): Radicals with the same radicand combined like terms
\[ 2\sqrt{2} + \sqrt{5} - 6\sqrt{2} = -4\sqrt{2} + \sqrt{5} \]
Verify: check numerically
Why: The original is about 2.83 plus 2.24 minus 8.49, which is about negative 3.42, and the answer is about negative 5.66 plus 2.24, also about negative 3.42. The two unlike terms genuinely cannot be combined further.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698
Sorting
Only identical radicands.
Sort into buckets
Sort each pair by whether the two terms combine.
Every combining pair shares a radicand exactly. The coefficients are irrelevant to whether combining is possible — only to what the result is.
Worked example
Guided Practice 1 to 3.
\[ \text{Simplify } \sqrt{3} + 2\sqrt{3}, \; 3\sqrt{5} - 2\sqrt{5} \text{ and } \sqrt{7} + \sqrt{2} + 3\sqrt{7}. \]
Take the first
Why: One plus two.
\[ 3\sqrt{3} \]
Take the second
Why: Three minus two.
\[ \sqrt{5} \]
Group the third
Why: Two have root seven.
\[ (1 + 3)\sqrt{7} + \sqrt{2} \]
Finish the third
Why: Four of them.
\[ 4\sqrt{7} + \sqrt{2} \]
Figure (svg): Radicals with the same radicand combined like terms
\[ 3\sqrt{3}, \quad \sqrt{5}, \quad 4\sqrt{7} + \sqrt{2} \]
Verify: notice the invisible coefficient
Why: A radical written alone has a coefficient of one, so root three plus two root three is one plus two of them. Forgetting that invisible one is the commonest slip in these.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698
Trap
\[ \sqrt{2} + \sqrt{5} = \sqrt{7} \]
Add what is under the radicals
Why: The radicals are being added, so their contents were added.
Root two is about 1.41 and root five about 2.24, totalling about 3.65, while root seven is about 2.65. This is the same error as splitting a radical over a sum in Lesson 9.3, seen from the other side.
\[ \sqrt{2} + \sqrt{5} \text{ cannot be combined} \]
Combine only when the radicands are identical
Why: Different radicands stay as separate terms.
An answer with two radical terms in it is usually finished rather than unfinished.
Faded example
The radicand does not change.
Fill in the blanks
2\sqrt2 - 6 - 6\sqrt-4 = (___)\sqrt___ = ___\sqrt___
Why: The distributive property factors the common radical out, leaving the coefficients to combine. The root two is carried through unchanged.
Elimination
Adding two radicals.
Eliminate the wrong options
What is root 2 plus 3 root 2?
Survives elimination: A
Why: One of them plus three of them is four of them, with the radical untouched. Every wrong option changes the radicand, which addition never does.
Socratic
It is a specific number.
Discussion prompt
Explain why root two can be treated as though it were a variable when combining terms. Then say what would be lost by evaluating it first.
Hint: What does the distributive property require?
Answer:
The distributive property works for any quantity, whether or not its value is known — two of something minus six of something is negative four of that something regardless of what it is. Root two is a fixed number, so the property certainly applies to it.
Evaluating it first would replace an exact quantity with a rounded decimal, so the answer would be approximate and could not be checked exactly. Keeping the radical means negative four root two is exact, and it also makes it visible which terms are alike — a decimal hides that entirely.
Section
Section 2
Concept
Two radicals with different radicands may become like radicals once each is simplified. Always remove perfect square factors before deciding that terms cannot be combined.
Root twenty-seven is three root three in disguise.
Figure (svg): Radicals simplified before they can be combined
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698 — Example 1, part b, and its Look Back to the simplifying of Lesson 9.3
Picture it
Simplify, then combine.
Figure (svg): Radicals simplified before they can be combined
Reported without simplifying, the expression would have looked like two unlike terms that could not be combined. The whole answer depends on doing Lesson 9.3's work first.
Worked example
This is Example 1, part b, from the textbook.
\[ \text{Simplify } 4\sqrt{3} - \sqrt{27}. \]
Look for a square factor
Why: Twenty-seven is nine times three.
\[ \sqrt{9 \cdot 3} \]
Apply the product property
Why: Split the radical.
\[ 3\sqrt{3} \]
Now the radicands match
Why: Both are root three.
\[ 4\sqrt{3} - 3\sqrt{3} \]
Combine
Why: Four minus three.
\[ \sqrt{3} \]
Figure (svg): Radicals simplified before they can be combined
\[ 4\sqrt{3} - \sqrt{27} = \sqrt{3} \]
Verify: check numerically
Why: Four times 1.732 is about 6.93, and the root of twenty-seven is about 5.20, whose difference is about 1.73 — which is the root of three. The simplification turned two unlike terms into one clean answer.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698
Faded example
A square factor is hiding.
Fill in the blanks
\sqrt9 = \sqrt3 \cdot 3} = ___\sqrt___
Why: Twenty-seven is nine times three, so its root is three root three. That makes it a like radical with root three, which it did not appear to be.
Worked example
Guided Practice 4 to 6.
\[ \text{Simplify } \sqrt{8} + \sqrt{2}, \; \sqrt{18} - \sqrt{2} \text{ and } 5\sqrt{3} - \sqrt{12}. \]
Take the first
Why: Eight is four times two.
\[ 2\sqrt{2} + \sqrt{2} = 3\sqrt{2} \]
Take the second
Why: Eighteen is nine times two.
\[ 3\sqrt{2} - \sqrt{2} = 2\sqrt{2} \]
Take the third
Why: Twelve is four times three.
\[ 5\sqrt{3} - 2\sqrt{3} \]
Finish the third
Why: Five minus two.
\[ 3\sqrt{3} \]
Figure (svg): Radicals simplified before they can be combined
\[ 3\sqrt{2}, \quad 2\sqrt{2}, \quad 3\sqrt{3} \]
Verify: notice what they had in common
Why: In every case one radicand was a perfect square times the other, so simplifying made them match. Radicands that differ by a square factor always turn out to be like radicals.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-698
Error analysis
The student was asked to simplify four root three minus the root of twenty-seven.
Annotate
On: \( \begin{aligned} 4\sqrt{3} - \sqrt{27} &= \text{unlike radicands} \\ &\text{so it cannot be simplified} \end{aligned} \)
Declaring an expression unsimplifiable is a real answer and must therefore be justified, not assumed. Simplifying every radical first is what makes such a declaration trustworthy.
Sorting
Simplify each first.
Sort into buckets
Sort each pair by whether they combine after simplifying.
Four of the six pairs combine and none of them looked as though it would. Simplifying first is what turns a wrong answer of cannot be combined into the right one.
Hypothesis
After both are simplified.
Predict first
What relationship between the radicands makes this happen?
Correct: One is a perfect square times the other.
\[ 27 = 9 \cdot 3, \quad 18 = 9 \cdot 2, \quad 12 = 4 \cdot 3 \]
Why: Twenty-seven is nine times three and eighteen is nine times two, so in each case removing the square factor leaves the same radicand. Being larger or even is irrelevant, and differing by a square is a different condition entirely — five and nine differ by four but do not simplify to like radicals. The test is to factor each radicand and see whether the square-free parts match.
Socratic
The radicands look different.
Discussion prompt
Explain why an expression should be fully simplified before declaring that its terms cannot be combined. Then say what such a declaration actually claims.
Hint: What might simplifying reveal?
Answer:
Simplifying can change a radicand, so two terms that look unlike may become alike — root twenty-seven becomes three root three, and suddenly it matches. Judging before simplifying is judging on appearances that are about to change.
The declaration claims that no further combining is possible, which is a statement about every form the expression could take rather than about the form it happens to be written in. Only after every radical is in simplest form is such a claim supportable.
Section
Section 3
Concept
Radicals multiply by combining their radicands. When one factor is a sum, the distributive property applies exactly as it does for polynomials.
\[ \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \]
The product often simplifies to a whole number.
Figure (svg): Three ways radicals are multiplied
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 699-699 — Example 2, Multiply Radicals
Picture it
Two radicals, a bracket, a pair.
Figure (svg): Three ways radicals are multiplied
The third line is the important one: multiplying a sum by the matching difference removes the radical completely, which the next section puts to work.
Worked example
This is Example 2 from the textbook.
\[ \text{Simplify } \sqrt{2} \cdot \sqrt{8}, \; \sqrt{2}(5 + \sqrt{3}) \text{ and } (2 + \sqrt{3})(2 - \sqrt{3}). \]
Take the first
Why: Two times eight is sixteen.
\[ \sqrt{16} = 4 \]
Distribute the second
Why: Root two across the bracket.
\[ 5\sqrt{2} + \sqrt{6} \]
Recognise the third
Why: A sum and a difference.
\[ 2^2 - (\sqrt{3})^2 \]
Evaluate it
Why: Four minus three.
\[ 1 \]
Figure (svg): Three ways radicals are multiplied
\[ 4, \quad 5\sqrt{2} + \sqrt{6}, \quad 1 \]
Verify: check the third numerically
Why: Two plus 1.732 is about 3.73 and two minus 1.732 is about 0.268, whose product is about one. The radical really does vanish, which is worth confirming the first time you meet it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 699-699
Faded example
Both terms.
Fill in the blanks
\sqrt5(5 + \sqrt6) = ___\sqrt___ + \sqrt___}
Why: The first product keeps the radical and gains a coefficient; the second combines two radicands into one. Both terms of the bracket must be reached.
Worked example
Guided Practice 7 to 9.
\[ \text{Simplify } \sqrt{3} \cdot \sqrt{12}, \; \sqrt{5}(\sqrt{2} + 1) \text{ and } (\sqrt{2} + 1)(\sqrt{2} - 1). \]
Take the first
Why: Three times twelve is thirty-six.
\[ 6 \]
Distribute the second
Why: Root five across the bracket.
\[ \sqrt{10} + \sqrt{5} \]
Recognise the third
Why: A sum and a difference.
\[ (\sqrt{2})^2 - 1^2 \]
Evaluate it
Why: Two minus one.
\[ 1 \]
Figure (svg): Three ways radicals are multiplied
\[ 6, \quad \sqrt{10} + \sqrt{5}, \quad 1 \]
Verify: note when a product is rational
Why: The first and third gave whole numbers and the second did not. A product of radicals is rational exactly when the combined radicand is a perfect square, or when the sum and difference pattern applies.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 699-699
Trap
\[ \sqrt{2}(5 + \sqrt{3}) = 5\sqrt{2} + \sqrt{3} \]
Multiply the first term and copy the second
Why: The root three already had a radical, so it looked done.
Both terms in the bracket must be multiplied by root two, so the second becomes root two times root three, which is root six. Leaving it as root three loses a factor entirely.
\[ = 5\sqrt{2} + \sqrt{6} \]
Multiply every term inside the bracket
Why: The distributive property has no exceptions.
Checking numerically catches this: the correct value is about 9.52 and the wrong one about 8.80.
Translation
Combine radicands or distribute.
Match the pairs
Why: Three of the four came out rational, which happens more often with radicals than with variables. The one that did not had a bracket whose terms were unlike.
Elimination
Some products are rational.
Eliminate the wrong options
Which of these simplifies to a rational number?
Survives elimination: A
Why: The sum and difference pattern makes the cross terms cancel and squares away the radical, leaving four minus three. That is exactly what the next section exploits.
Socratic
It is the pattern from Lesson 10.3.
Discussion prompt
Explain why multiplying a sum by the matching difference removes the radical. Then say what would happen if the signs matched instead.
Hint: What happens to the cross products, and to the square?
Answer:
The two cross products are opposites and cancel, so nothing containing a single radical survives. What is left is the first term squared minus the radical squared, and squaring a square root removes it — root three squared is simply three.
With matching signs the cross products would add rather than cancel, giving a middle term of four root three in the case above. So squaring a sum keeps the radical and multiplying by the difference removes it, which is why only one of the two is useful for rationalising.
Section
Section 4
Concept
A denominator that is a single radical is cleared by multiplying by that radical over itself. A denominator that is a sum containing a radical is cleared by multiplying by the matching difference.
The second case uses the sum and difference pattern.
Figure (svg): Two ways of clearing a radical from a denominator
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 699-699 — Example 3, Simplify Radicals, and its Study Tip on multiplying by one
Picture it
Match the denominator.
Figure (svg): Two ways of clearing a radical from a denominator
Both multipliers equal one, so neither changes the value of the expression. Choosing which one is the only decision to make.
Worked example
This is Example 3, part a, from the textbook.
\[ \text{Simplify } \dfrac{3}{\sqrt{5}}. \]
Choose the multiplier
Why: The radical over itself.
\[ \tfrac{\sqrt{5}}{\sqrt{5}} \]
Multiply
Why: Numerators and denominators.
\[ \tfrac{3\sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} \]
Simplify the denominator
Why: A root times itself.
\[ 5 \]
Write the answer
Why: The radical is now on top.
\[ \tfrac{3\sqrt{5}}{5} \]
Figure (svg): Two ways of clearing a radical from a denominator
\[ \dfrac{3}{\sqrt{5}} = \dfrac{3\sqrt{5}}{5} \]
Verify: compare decimals
Why: Three divided by 2.236 is about 1.342, and three times 2.236 divided by five is also about 1.342. Multiplying by one changed the form and not the value.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 699-699
Faded example
Flip the sign, keep the terms.
Fill in the blanks
\text- 2 + \sqrt1, \text___ \dfrac___} \sqrt___}___}, \text___ 4 - 3 = ___
Why: Only the sign between the terms changes; the terms themselves stay the same. The resulting denominator is the first term squared minus the radicand.
Worked example
This is Example 3, part b, from the textbook.
\[ \text{Simplify } \dfrac{1}{2 + \sqrt{3}}. \]
Choose the multiplier
Why: The matching difference.
\[ \tfrac{2 - \sqrt{3}}{2 - \sqrt{3}} \]
Multiply
Why: The denominator uses the pattern.
\[ \tfrac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \]
Evaluate the denominator
Why: Four minus three.
\[ 1 \]
Write the answer
Why: Nothing left underneath.
\[ 2 - \sqrt{3} \]
Figure (svg): A sum and difference of radicals producing a rational result
\[ \dfrac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} \]
Verify: compare decimals
Why: One divided by 3.732 is about 0.268, and two minus 1.732 is also about 0.268. The denominator came out as exactly one, which is why the answer has no fraction at all.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 699-699
Trap
\[ \dfrac{1}{2 + \sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{2\sqrt{3} + 3} \]
Multiply by the radical over itself, as before
Why: The single-radical method was applied to a sum.
The denominator still contains a radical, so nothing has been achieved. A sum needs the matching difference, which makes the cross terms cancel.
\[ \cdot \dfrac{2 - \sqrt{3}}{2 - \sqrt{3}} = 2 - \sqrt{3} \]
Match the multiplier to the shape of the denominator
Why: Single radical or sum.
Checking that the new denominator has no radical is the test of whether the right multiplier was chosen.
Matching
Single radical or sum.
Match the pairs
Why: A lone radical is multiplied by itself and a sum by the matching difference. The last row flips a minus to a plus, which is the same rule applied the other way.
Faded example
The denominator becomes rational.
Fill in the blanks
\dfrac32} \cdot \dfrac___}___} = \dfrac___)}___}} = \dfrac___}___}
Why: The denominator becomes twenty-two, which shares a factor of eleven with the numerator. Simplifying afterwards is often possible and worth checking for.
Socratic
A calculator handles either form.
Discussion prompt
Give a reason for rationalising beyond the historical one of easier hand division. Then say where this particular pattern reappears.
Hint: Think about comparing two answers.
Answer:
It gives a single standard form, so two people can compare answers without checking whether differently written expressions are secretly equal. One over two plus root three and two minus root three look nothing alike, and knowing they are the same number is not obvious without doing the work.
The same sum-and-difference multiplier reappears wherever a radical must be removed from a denominator, including in later courses when working with complex numbers. The pattern is more general than this one use, which is why it is worth recognising rather than merely following.
Section
Section 5
Concept
When two values of a radical model share a radicand after simplifying, their difference is exact rather than a rounded decimal. Simplifying before subtracting is what reveals this.
The answer is often a single radical.
Figure (svg): Distance to the horizon against eye-level height
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 700-700 — Example 4, Use a Radical Model, on the distance to the horizon
Picture it
The difference comes out exactly.
Figure (svg): Distance to the horizon against eye-level height
Both distances turned out to be multiples of root three, so their difference is a single radical. Rounding first would have given about 1.73 and hidden that.
Worked example
This is Example 4 from the textbook.
\[ \text{With } d = \sqrt{\tfrac{3h}{2}}, \text{ compare eye heights of } 32 \text{ and } 18 \text{ feet.} \]
Evaluate at thirty-two
Why: Three times thirty-two over two.
\[ \sqrt{48} \]
Simplify it
Why: Sixteen times three.
\[ 4\sqrt{3} \]
Evaluate at eighteen
Why: Three times eighteen over two.
\[ \sqrt{27} = 3\sqrt{3} \]
Subtract
Why: Like radicals.
\[ \sqrt{3} \]
Figure (svg): Distance to the horizon against eye-level height
\[ 4\sqrt{3} - 3\sqrt{3} = \sqrt{3} \approx 1.7 \text{ miles} \]
Verify: check the two distances
Why: Four root three is about 6.93 miles and three root three about 5.20, differing by about 1.73 — the root of three. The exact answer is a single radical, which no decimal calculation would have revealed.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 700-700
Faded example
The radicands match.
Fill in the blanks
\sqrt4 = 1\sqrt___, \quad \sqrt___ = 3\sqrt___ \;\Longrightarrow\; \text___ = ___\sqrt___
Why: Both radicands were multiples of three by perfect squares, so both simplified to multiples of root three. The difference is one of them, which is simply root three.
Worked example
Reading the flattening from the radical.
\[ \text{How much higher must you be to see twice as far?} \]
Note the relationship
Why: Distance depends on the root of height.
\[ d \propto \sqrt{h} \]
Ask for double the distance
Why: The root must double.
\[ \sqrt{h} \times 2 \]
Find the height factor
Why: Squaring the doubling.
\[ h \times 4 \]
State it
Why: Four times the height.
\[ \times 4 \]
Figure (svg): Distance to the horizon against eye-level height
\[ h \times 4 \;\Longrightarrow\; d \times 2 \]
Verify: test with the numbers
Why: At eight feet the distance is the root of twelve, about 3.46 miles, and at thirty-two feet it is about 6.93 — twice as far for four times the height. That is the same flattening met in Lesson 12.1.
Trap
\[ 6.93 - 5.20 = 1.73 \text{ miles} \]
Evaluate both distances as decimals, then subtract
Why: The question asks how much farther, which is a number.
The answer is right to two decimal places but hides that the exact difference is root three. Worse, rounding each distance first means the difference carries both roundings and its last digit is not reliable.
\[ 4\sqrt{3} - 3\sqrt{3} = \sqrt{3} \approx 1.7 \]
Simplify and subtract exactly, then round once at the end
Why: Exact work first, rounding last.
This is the same discipline as in Lesson 9.3, where rounding partway through was the trap.
Prediction
The distance depends on the root of the height.
Predict first
To see twice as far, how much higher must you be?
Correct: Four times as high.
\[ h = 8: \; d \approx 3.46 \qquad h = 32: \; d \approx 6.93 \]
Why: The distance is proportional to the square root of the height, so doubling the distance requires the root to double, which requires the height to quadruple. Going from eight feet to thirty-two feet takes the view from about 3.5 miles to about 6.9. That is the diminishing return characteristic of every square root model, and it is why crow's nests are placed as high as they can practically be.
Sorting
Which form does each question want?
Sort into buckets
Sort each request by the form of answer it calls for.
Example 4 asks for both, which is common: an expression first and then its value. Giving only one of the two answers half the question.
Socratic
A sailor wants a number of miles.
Discussion prompt
Say what the exact form of the answer adds beyond the decimal. Then say when the decimal is the right thing to report.
Hint: What does root three tell you that 1.73 does not?
Answer:
The exact form shows the structure: the difference is exactly one unit of the same radical that both distances are built from, which says the two heights are related in a particular way. It is also checkable, since squaring it recovers three, and usable in further calculation without accumulating rounding error.
The decimal is right when someone is going to act on the number — deciding whether the extra view is worth climbing for, say. Then about 1.7 miles is what matters and the radical is of no practical use. Reporting both, with the exact value in the working, is the usual convention.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Operation | Requires | Effect on the radicand |
|---|---|---|
| Adding | matching radicands | unchanged; only coefficients combine |
| Multiplying | nothing; any radicals may be multiplied | the radicands multiply together |
| Rationalising | a radical in a denominator | the denominator's radical is removed |
Adding is the fussy one, because it demands the radicands already match. Multiplying works on anything, which is why it is used to fix denominators.
Pattern
To simplify any expression involving radicals, these five moves cover it.
Step one comes first because it can turn unlike radicands into like ones, and that changes what step three is able to combine.
OpenStax Elementary Algebra 2e, §9.3 Add and Subtract Square Roots §9.3
Check
Only matching radicands.
Check your understanding
Simplify 2 root 2 + root 5 - 6 root 2.
Answer: A
Why: The two root two terms combine to negative four root two, and root five has no partner so it stays as a separate term.
Check
Simplify first.
Check your understanding
Simplify 4 root 3 - root 27.
Answer: A
Why: The root of twenty-seven is three root three, so the expression becomes four root three minus three root three, which is root three.
Check
Match the multiplier to the denominator.
Check your understanding
Simplify 1 over (2 + root 3).
Answer: A
Why: Multiplying by two minus root three over itself gives a denominator of four minus three, which is one, leaving two minus root three.
Real world
This is the schooner question from the lesson opener. The distance d in miles you can see to the horizon is modelled by the square root of three h over two, where h is your eye-level height in feet.
Discussion prompt
Your eye-level height is 32 feet and your friend's is 18. Write an expression for how much farther you can see, simplify it, and say how much higher you would need to be to see twice as far as your friend.
Hint: Simplify each distance before subtracting.
Answer:
\[ \sqrt{48} - \sqrt{27} = 4\sqrt{3} - 3\sqrt{3} = \sqrt{3} \approx 1.7 \text{ miles} \]
Both distances turned out to be multiples of root three, so the difference is exactly root three miles rather than a messy decimal.
To see twice as far as your friend you would need twice their distance, which means four times their height — seventy-two feet rather than eighteen. That is well above the thirty-two feet you are already at, which shows how quickly the returns diminish: your fourteen extra feet bought only 1.7 miles, and the next doubling would cost forty more.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Simplify 4 root 3 minus the root of 27.
Correct: root 3.
\[ 4\sqrt{3} - \sqrt{27} = 4\sqrt{3} - 3\sqrt{3} = \sqrt{3} \]
Why: Twenty-seven contains a perfect square factor of nine, so its root is three root three — which makes it a like radical with four root three, and the difference is one of them. The first option is the trap: the radicands do differ as written, but declaring an expression unsimplifiable is a claim about every form it could take, not about the form it happens to be in, so simplifying must come first. Checking numerically settles it, since four times 1.732 minus 5.196 is about 1.73, which is root three rather than anything unsimplifiable. The third option subtracts the radicands, and the fourth treats the twenty-seven as a coefficient.
Explain it
They wrote that the root of two plus the root of five is the root of seven.
Discussion prompt
In no more than four sentences, explain why radicands are not added. Then give them a check that takes ten seconds.
Hint: Think of the radicals as objects to be counted.
Answer:
A usable answer: adding radicals is like adding terms in algebra — two of a thing plus three of the same thing is five of it, but a root two and a root five are different things and cannot be pooled. What adds is how many of each you have, never what is inside them.
The check is to work both out roughly. Root two is about 1.4 and root five about 2.2, adding to about 3.6, while root seven is only about 2.6. They are plainly different numbers.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Combining is fixed by checking that the radicands are identical, exactly as with like terms. Simplifying first is fixed by refusing to declare anything unsimplifiable until every radical is in simplest form. Distributing is fixed by multiplying every term inside the bracket. The multiplier is fixed by asking whether the denominator is a lone radical or a sum. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write a sum of radicals with three terms, two of which share a radicand, and combine them, marking clearly which coefficients merged and which term stood alone. Underneath, take two radicals that look unlike, simplify both to reveal a match, and combine them — then write beside it the numerical check that confirms the answer. In the middle, work three products: two radicals multiplied directly, a radical distributed across a bracket, and a sum times its matching difference, noting for the last one that the radical vanishes and why. Beneath that, rationalise two denominators, one a lone radical and one a sum, showing the multiplier you chose in each case and verifying that the new denominator has no radical. In the lower corner, evaluate a radical model at two inputs, simplify both results, subtract them exactly, and only then round — writing the exact and rounded answers side by side. Finally, in the margin, write which operation requires matching radicands and which do not.
Every answer on your page should survive a rough numerical check to one decimal place. If one does not, the likeliest cause is a radicand that was combined when only the coefficients should have been.
Recap
Five things, and the second is what makes the first trustworthy.
| If the question says | Your first move is |
|---|---|
| Add or subtract radicals | Simplify each, then match radicands |
| The radicands look different | Simplify before concluding anything |
| Multiply a radical by a bracket | Distribute to every term |
| A lone radical is in the denominator | Multiply by that radical over itself |
| A sum is in the denominator | Multiply by the matching difference |
Lesson 12.3 solves equations containing radicals. Squaring both sides clears the radical, and — exactly as in Lesson 11.7 — that step can produce a candidate the original equation never allowed.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.2 Operations with Radical Expressions §12.2, pp. 698-703 — everything on these slides traces back here
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