Writing radical expressions in simplest form. Includes the three conditions for simplest form, the product property of radicals and removing perfect square factors, choosing an efficient factorisation, the quotient property, rationalising a denominator, and applying the whole procedure to a boat speed model.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 9 — Quadratic Equations and Functions
Simplifying Radicals
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. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-516 — the lesson these objectives are drawn from
Warm-up
Lesson 9.2 produced answers like the root of forty-eight. This lesson makes such answers presentable.
Discussion prompt
The root of forty-eight and four times the root of three are the same number. Check that with a calculator, then say which form you would rather compare with the root of seventy-five.
Hint: Work out both to two decimal places.
Answer:
\[ \sqrt{48} \approx 6.93 \quad \text{and} \quad 4\sqrt{3} \approx 4(1.732) = 6.93 \]
They agree, as they must. The second form is easier to compare with other radicals, because the root of seventy-five is five root three and two multiples of the same radical can be compared at a glance — which is exactly what Lesson 12.2 will need.
Concept
A radical expression is in simplest form when it has no perfect square factors other than one in the radicand, no fractions in the radicand, and no radicals in the denominator of a fraction.
simplest form of a radical expression — A radical expression with no perfect square factors other than one in the radicand, no fractions in the radicand, and no radicals in any denominator.
Two properties of radicals do all the work of getting there.
Figure (svg): The three conditions for a radical expression to be in simplest form
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511
Section
Section 1
Concept
The root of a product equals the product of the roots, provided both factors are non-negative. That is what allows a perfect square factor to be pulled out of a radicand.
\[ \sqrt{ab} = \sqrt{a} \cdot \sqrt{b}, \quad a \ge 0, \; b \ge 0 \]
It splits over multiplication and never over addition.
Figure (svg): The product property of radicals stated and illustrated
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511 — the Product Property of Radicals and its worked illustration
Picture it
Multiplication yes, addition no.
Figure (svg): The product property of radicals stated and illustrated
The counterexample underneath is the one to memorise. The root of a sum is almost never the sum of the roots, and the one place it looks plausible is exactly where it fails.
Worked example
This is Example 1, part a, from the textbook.
\[ \text{Simplify } \sqrt{50}. \]
Find a perfect square factor
Why: Fifty is twenty-five times two.
\[ \sqrt{25 \cdot 2} \]
Apply the product property
Why: Split the radical.
\[ \sqrt{25} \cdot \sqrt{2} \]
Evaluate the perfect square
Why: The root of twenty-five is five.
\[ 5\sqrt{2} \]
Check the remaining radicand
Why: Two has no square factors.
Figure (svg): Pulling a perfect square factor out of a radicand
\[ \sqrt{50} = 5\sqrt{2} \]
Verify: compare the decimals
Why: The root of fifty is about 7.071, and five times 1.4142 is also about 7.071. Simplifying never changes the value, so any simplification can be checked by evaluating both forms.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511
Matching
Pull out the perfect square.
Match the pairs
Why: Every one of these leaves a prime under the radical, which is the sign that the job is done. The whole number in front is always the root of the perfect square factor that was removed.
Worked example
Guided Practice 1 to 4.
\[ \text{Simplify } \sqrt{12}, \; \sqrt{32}, \; \sqrt{75} \text{ and } \sqrt{180}. \]
Take twelve
Why: Four times three.
\[ 2\sqrt{3} \]
Take thirty-two
Why: Sixteen times two.
\[ 4\sqrt{2} \]
Take seventy-five
Why: Twenty-five times three.
\[ 5\sqrt{3} \]
Take a hundred and eighty
Why: Thirty-six times five.
\[ 6\sqrt{5} \]
Figure (svg): Pulling a perfect square factor out of a radicand
\[ 2\sqrt{3}, \quad 4\sqrt{2}, \quad 5\sqrt{3}, \quad 6\sqrt{5} \]
Verify: check that no radicand can still be reduced
Why: Three, two, three and five are all prime, so none of them contains a perfect square factor. Checking the final radicand is the step that decides whether the expression is genuinely in simplest form.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511
Trap
\[ \sqrt{25 + 4} = \sqrt{25} + \sqrt{4} = 5 + 2 = 7 \]
Apply the product property to a sum
Why: The rule splits a radical, so it looks as though it should split this one.
Twenty-five plus four is twenty-nine, whose root is about 5.39, not seven. The property is stated for a product and says nothing at all about a sum.
\[ \sqrt{25 + 4} = \sqrt{29} \approx 5.39 \]
Add underneath the bar and stop, since twenty-nine has no square factors
Why: Nothing further can be done.
A radical of a sum usually cannot be simplified at all, and recognising that is a legitimate finish.
Faded example
Perfect square first.
Fill in the blanks
\sqrt25 = \sqrt5 \cdot 3} = \sqrt___ \cdot \sqrt___ = ___\sqrt___
Why: Twenty-five is the largest perfect square dividing seventy-five, and three is what remains. Using a smaller square factor would still work but would need a second pass.
Elimination
Simplifying the root of forty.
Eliminate the wrong options
Which move is correct?
Survives elimination: A
Why: Only the multiplicative split is valid, giving two root ten. Options B and D are the same error dressed differently, and comparing decimals exposes both in a few seconds.
Socratic
It fails for sums, so why not for products?
Discussion prompt
Explain why the root of a product equals the product of the roots. Then say why the same reasoning does not work for a sum.
Hint: Square both sides and see what happens.
Answer:
Squaring the product of the two roots multiplies each of them by itself, and since multiplication can be rearranged freely, the result is a times b. So the product of the roots is a number whose square is a times b, which is exactly what the root of the product means — and both are non-negative, so they are the same number.
For a sum the same move fails. Squaring the sum of two roots gives a plus b plus twice the product of the roots, and that extra middle term is not nought unless one of the numbers is. It is the same middle term that will appear in Lesson 10.3, and it is precisely what makes squaring a sum different from squaring a product.
Section
Section 2
Concept
A radicand can often be factored in more than one way. Any perfect square factor works, but the largest one finishes the job in a single pass.
The answer is the same either way.
Figure (svg): Simplifying the same radical by two different factorisations
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511 — Example 1, part b, and its Study Tip on efficient factoring
Picture it
Four times twelve, or sixteen times three.
Figure (svg): Simplifying the same radical by two different factorisations
The short route needs you to notice that sixteen divides forty-eight. Knowing the perfect squares by sight is what turns the long route into the short one.
Worked example
This is Example 1, part b, from the textbook.
\[ \text{Simplify } \sqrt{48} \text{ using the factor } 4. \]
Factor out four
Why: Forty-eight is four times twelve.
\[ \sqrt{4 \cdot 12} \]
Factor again
Why: Twelve is four times three.
\[ \sqrt{4 \cdot 4 \cdot 3} \]
Apply the property
Why: Two fours make sixteen.
\[ \sqrt{16} \cdot \sqrt{3} \]
Simplify
Why: The root of sixteen is four.
\[ 4\sqrt{3} \]
Figure (svg): Simplifying the same radical by two different factorisations
\[ \sqrt{48} = 4\sqrt{3} \]
Verify: check the radicand is finished
Why: Three is prime, so no square factor remains and the expression is in simplest form. Had the work stopped at two times the root of twelve, twelve still contains a factor of four and the job would be half done.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511
Sorting
Check the radicand for square factors.
Sort into buckets
Sort each expression by whether it is in simplest form.
The three unfinished ones hide a four or a nine. Glancing at the radicand for a factor of four, nine, sixteen or twenty-five catches nearly every case.
Worked example
The Study Tip's route.
\[ \text{Simplify } \sqrt{48} \text{ using the largest perfect square factor.} \]
Find the largest square factor
Why: Sixteen divides forty-eight.
\[ 48 = 16 \cdot 3 \]
Apply the property
Why: Split the radical.
\[ \sqrt{16} \cdot \sqrt{3} \]
Simplify
Why: The root of sixteen is four.
\[ 4\sqrt{3} \]
Compare with the long route
Why: Same answer, two steps fewer.
Figure (svg): Simplifying the same radical by two different factorisations
\[ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \]
Verify: confirm both routes agree
Why: Both give four root three, about 6.93, and the root of forty-eight is about 6.93. Choosing a factorisation is a matter of efficiency and never of correctness, which is worth knowing when a smaller factor is all you can spot.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-511
Error analysis
The student was simplifying the root of seventy-two.
Annotate
On: \( \begin{aligned} \sqrt{72} &= \sqrt{4 \cdot 18} \\ &= 2\sqrt{18} \\ \text{so the answer is } &2\sqrt{18} \end{aligned} \)
The answer was not wrong, only unfinished, and that is a distinction worth being precise about. Simplest form is a stated standard with three conditions, and an expression that fails the first one has not met it however correct its value.
Faded example
Another square is still inside.
Fill in the blanks
\sqrt3 = 2\sqrt6 = 2\sqrt___ = 2 \cdot ___\sqrt___ = ___\sqrt___
Why: Stopping at two root eighteen gives the right value in the wrong form. Using thirty-six from the start would have reached six root two immediately, but either route ends in the same place.
Translation
Name the biggest square that divides it.
Match the pairs
Why: Two of these share a largest square factor of thirty-six but leave different numbers behind — two and five respectively. What comes out and what stays in are separate questions.
Socratic
Simplifying could go on forever.
Discussion prompt
Describe a test that tells you a radicand cannot be simplified further. Then say why a prime radicand always passes it.
Hint: Ask what factors the radicand has.
Answer:
Factor the radicand into primes and look for any prime appearing twice. A repeated prime is a perfect square hiding inside, and it can be pulled out; if no prime repeats, then the only perfect square factor is one and the expression is finished.
A prime radicand has just itself and one as factors, so no prime can repeat and the test passes immediately. That is why the answers in this lesson so often end with the root of two, three or five — those are what survive after every repeated factor has been removed.
Section
Section 3
Concept
The root of a quotient equals the quotient of the roots, provided the numerator is non-negative and the denominator is strictly positive. This removes a fraction from underneath a radical.
\[ \sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}, \quad a \ge 0, \; b > 0 \]
The denominator must be positive, not merely non-negative.
Figure (svg): The quotient property of radicals stated and illustrated
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 512-512 — the Quotient Property of Radicals and Example 2
Picture it
Numerator and denominator separately.
Figure (svg): The quotient property of radicals stated and illustrated
Reducing the fraction before splitting usually saves effort, because a simpler fraction often turns out to be a ratio of two perfect squares.
Worked example
This is Example 2 from the textbook.
\[ \text{Simplify } \sqrt{\dfrac{320}{500}}. \]
Reduce the fraction first
Why: Divide out a common factor of twenty.
\[ \sqrt{\tfrac{16}{25}} \]
Apply the quotient property
Why: Split into two radicals.
\[ \dfrac{\sqrt{16}}{\sqrt{25}} \]
Evaluate both
Why: Both are perfect squares.
\[ \dfrac{4}{5} \]
Check the form
Why: No radical is left at all.
Figure (svg): The quotient property of radicals stated and illustrated
\[ \sqrt{\dfrac{320}{500}} = \dfrac{4}{5} \]
Verify: square the answer
Why: Four fifths squared is sixteen twenty-fifths, which is 0.64, and 320 divided by 500 is also 0.64. The radical vanished entirely because the reduced fraction was a ratio of two perfect squares.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 512-512
Faded example
The fraction gets simpler first.
Fill in the blanks
\sqrt255} = \sqrt______}} = \dfrac______}
Why: Reducing turned an ugly fraction into a ratio of two perfect squares, so the radical disappeared completely. That will not always happen, but it is worth looking for before doing anything else.
Worked example
Guided Practice 8, and the summary's second example.
\[ \text{Simplify } \sqrt{\dfrac{75}{12}} \text{ and } \sqrt{\dfrac{5}{16}}. \]
Reduce the first
Why: Divide top and bottom by three.
\[ \sqrt{\tfrac{25}{4}} \]
Split and evaluate
Why: Both are perfect squares.
\[ \tfrac{5}{2} \]
Split the second
Why: The numerator is not a perfect square.
\[ \dfrac{\sqrt{5}}{\sqrt{16}} \]
Evaluate the denominator
Why: Its root is four.
\[ \dfrac{\sqrt{5}}{4} \]
Figure (svg): The solution to Worked example when only one part is a perfect square shown as a ladder of expressions, one row per algebraic move
\[ \dfrac{5}{2}, \qquad \dfrac{\sqrt{5}}{4} \]
Verify: check the second condition of simplest form
Why: Neither answer has a fraction under a radical any more, and the second has its radical in the numerator where it is allowed. A radical in a numerator is perfectly acceptable; only a denominator is forbidden.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 512-513
Trap
\[ \sqrt{\tfrac{320}{500}} = \dfrac{\sqrt{320}}{\sqrt{500}} = \dfrac{8\sqrt{5}}{10\sqrt{5}} \]
Split first, then simplify each radical separately
Why: The property allows the split at any time.
It does, and this is not wrong — but it produces two messy radicals that then have to be cancelled. Reducing the fraction first gives sixteen over twenty-five and the radicals disappear.
\[ \sqrt{\tfrac{320}{500}} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5} \]
Reduce the fraction underneath the radical before splitting
Why: A simpler fraction is often a ratio of perfect squares.
Both routes reach four fifths, and one of them takes three lines instead of six.
Elimination
Check all three conditions.
Eliminate the wrong options
Which one still fails a condition of simplest form?
Survives elimination: A
Why: Reading option A as a radical whose radicand is a fraction, it fails the second condition and should be split into the root of five over four. The distinction between a radical of a fraction and a fraction of a radical is exactly what the quotient property manages.
Prediction
Some quotients simplify to a plain fraction.
Predict first
What has to be true for the root of a fraction to have no radical left?
Correct: Both numerator and denominator must be perfect squares, after reducing.
\[ \sqrt{\tfrac{16}{25}} = \tfrac{4}{5} \qquad \text{but} \qquad \sqrt{\tfrac{5}{16}} = \tfrac{\sqrt{5}}{4} \]
Why: Splitting produces one radical on top and one underneath, so both have to evaluate exactly for the radicals to disappear. Three hundred and twenty over five hundred worked because it reduced to sixteen over twenty-five; five over sixteen keeps its radical because five is not a perfect square. Reducing first matters because a fraction can hide this property until it is in lowest terms.
Socratic
The numerator is only required to be non-negative.
Discussion prompt
Explain why the quotient property demands b greater than nought rather than b at least nought. Then say what would go wrong at each of the two excluded values.
Hint: Consider b equal to nought and b negative separately.
Answer:
If b were nought the fraction itself would be undefined, and so would the root of the denominator on the right-hand side — division by nought is not permitted anywhere, so the statement could not hold. That is why nought has to be excluded even though it is a perfectly good radicand elsewhere.
If b were negative the root of b would be undefined over the reals, so the right-hand side would have no meaning even in cases where the left-hand side did — a negative over a negative is positive, so the fraction could be fine while its split version was not. Requiring b to be strictly positive rules out both failures at once.
Section
Section 4
Concept
A radical in a denominator is removed by multiplying the expression by a fraction equal to one, chosen so that the denominator becomes a perfect square. This is called rationalising the denominator.
\[ \dfrac{1}{\sqrt{7}} \cdot \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{7}}{7} \]
Multiplying by one changes the form and never the value.
Figure (svg): Removing a radical from a denominator
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 512-512 — Example 3, Rationalize the Denominator, and its Study Tip
Picture it
Split, simplify, multiply by one.
Figure (svg): Removing a radical from a denominator
The value is identical at every line. Only the writing changes, which is why a decimal check confirms the whole chain at once.
Worked example
This is Example 3 from the textbook.
\[ \text{Simplify } \sqrt{\dfrac{1}{18}}. \]
Apply the quotient property
Why: Split the radical.
\[ \dfrac{1}{\sqrt{18}} \]
Simplify the denominator
Why: Eighteen is nine times two.
\[ \dfrac{1}{3\sqrt{2}} \]
Multiply by a form of one
Why: Root two over root two.
\[ \dfrac{1}{3\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} \]
Simplify
Why: Three times two is six.
\[ \dfrac{\sqrt{2}}{6} \]
Figure (svg): Removing a radical from a denominator
\[ \sqrt{\dfrac{1}{18}} = \dfrac{\sqrt{2}}{6} \]
Verify: compare the decimals
Why: One over eighteen is about 0.0556, whose root is about 0.2357, and 1.4142 divided by six is also about 0.2357. Every line in the chain represents the same number written differently.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 512-512
Faded example
It has to clear the denominator's radical.
Fill in the blanks
\dfrac26} \cdot \dfrac___}}}___} = \dfrac___}___}
Why: Multiplying root two by itself gives two, and three times two is six. The form of one always matches the radical that needs clearing, which is what makes the choice mechanical rather than clever.
Worked example
The summary's third example.
\[ \text{Simplify } \dfrac{1}{\sqrt{7}} \text{ and } \sqrt{\dfrac{1}{3}}. \]
Multiply the first by one
Why: Root seven over root seven.
\[ \dfrac{\sqrt{7}}{\sqrt{7} \cdot \sqrt{7}} \]
Simplify
Why: A root times itself is the radicand.
\[ \dfrac{\sqrt{7}}{7} \]
Split the second
Why: Quotient property first.
\[ \dfrac{1}{\sqrt{3}} \]
Rationalise it
Why: Root three over root three.
\[ \dfrac{\sqrt{3}}{3} \]
Figure (svg): The solution to Worked example the simplest case shown as a ladder of expressions, one row per algebraic move
\[ \dfrac{\sqrt{7}}{7}, \qquad \dfrac{\sqrt{3}}{3} \]
Verify: notice the pattern
Why: In both cases the radical moved from the bottom to the top and the denominator became the old radicand. One over the root of any number equals the root of that number over the number itself, which is worth recognising on sight.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 513-513
Trap
\[ \dfrac{1}{\sqrt{7}} = \dfrac{1}{\sqrt{7} \cdot \sqrt{7}} = \dfrac{1}{7} \]
Multiply the denominator by root seven to clear the radical
Why: The goal was to remove the radical, and this removes it.
It also changes the value. One over the root of seven is about 0.378 and one seventh is about 0.143, so the two are nowhere near equal. The numerator has to be multiplied by the same thing.
\[ \dfrac{1}{\sqrt{7}} \cdot \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{7}}{7} \approx 0.378 \]
Multiply the whole fraction by a fraction equal to one
Why: Top and bottom both get the factor.
Multiplying by one is the only move that is guaranteed to leave a value alone.
Hypothesis
Root two over root two is used freely here.
Predict first
What does multiplying an expression by root two over root two do?
Correct: Changes how it is written but not its value.
\[ \dfrac{\sqrt{2}}{\sqrt{2}} = 1 \]
Why: Any non-zero quantity divided by itself is exactly one, and multiplying by one leaves a number alone. No rounding is involved because the exact radical is used rather than a decimal approximation, which is one more reason to keep radicals in exact form while working. The technique is the same as writing a half as two quarters — the value is untouched and only the appearance changes.
Elimination
Rationalising one over the root of five.
Eliminate the wrong options
Which form of one should be used?
Survives elimination: A
Why: The multiplier has to be the offending radical over itself, so that the denominator becomes a perfect square. Option C is the trap worth naming: it looks like the right numbers in the wrong arrangement and is not equal to one at all.
Socratic
Both forms are equally correct.
Discussion prompt
Give a reason for preferring a rational denominator. Then say whether that reason is as strong today as it once was.
Hint: Think about computing the decimal by hand.
Answer:
Dividing by an irrational number by hand is far harder than dividing by a whole one. Working out one divided by 1.4142 needs long division by a messy divisor, whereas the root of two over six needs only 1.4142 divided by six, which is easy — that convenience is where the convention came from.
With a calculator the practical advantage has largely gone, but the convention survives because it gives a single standard form, and a standard form lets two people compare answers without having to check whether their different-looking expressions are secretly equal. That is the same reason simplest form exists at all, and it matters more as expressions get more complicated.
Section
Section 5
Concept
When a model produces a radical, the exact simplified radical and the rounded decimal answer both matter. The radical is the answer and the decimal is what it means in practice.
The boat speed model is a quadratic with no linear term.
Figure (svg): A sailboat's maximum speed against its water line length
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 513-513 — Example 4, Simplify a Radical Expression, on boat speed
Picture it
Longer hull, faster boat.
Figure (svg): A sailboat's maximum speed against its water line length
Because speed depends on a square root, each extra foot of water line buys less speed than the one before. Designers meet that wall in practice, not just on paper.
Worked example
This is Example 4 from the textbook.
\[ \text{With } s^2 = \tfrac{16}{9}x, \text{ find the maximum speed for a } 32 \text{ foot water line.} \]
Substitute the length
Why: Thirty-two feet of water line.
\[ s^2 = \tfrac{16}{9}(32) \]
Take square roots
Why: Speed is positive.
\[ s = \sqrt{\tfrac{16}{9}} \cdot \sqrt{32} \]
Simplify each radical
Why: Four thirds, and four root two.
\[ \tfrac{4}{3} \cdot 4\sqrt{2} \]
Multiply
Why: Sixteen root two over three.
\[ \tfrac{16\sqrt{2}}{3} \]
Figure (svg): A sailboat's maximum speed against its water line length
\[ s = \dfrac{16\sqrt{2}}{3} \approx 7.5 \text{ knots} \]
Verify: square the answer back
Why: Sixteen root two over three, squared, is two hundred and fifty-six times two over nine, which is five hundred and twelve ninths — and sixteen ninths times thirty-two is also five hundred and twelve ninths. The exact form checks exactly, which a rounded one could not.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 513-513
Faded example
Both radicals simplify.
Fill in the blanks
s = \sqrt416} \cdot \sqrt___ = \tfrac______ \cdot ___\sqrt___ = \tfrac___\sqrt___}___
Why: The quotient property handles the fraction and the product property handles the thirty-two, and both are needed in the same problem. Multiplying the two whole-number parts at the end keeps the radical alone.
Worked example
Guided Practice 9.
\[ \text{Find the maximum speed for a } 50 \text{ foot water line.} \]
Substitute
Why: Fifty feet.
\[ s^2 = \tfrac{16}{9}(50) \]
Take roots
Why: Four thirds times the root of fifty.
\[ \tfrac{4}{3}\sqrt{50} \]
Simplify the radical
Why: Fifty is twenty-five times two.
\[ \tfrac{4}{3} \cdot 5\sqrt{2} \]
Multiply and round
Why: Twenty root two over three.
\[ \approx 9.4 \]
Figure (svg): A sailboat's maximum speed against its water line length
\[ s = \dfrac{20\sqrt{2}}{3} \approx 9.4 \text{ knots} \]
Verify: compare the two boats
Why: The water line grew by about fifty-six per cent and the speed by only about twenty-five per cent. A square root always flattens a change like that, which is exactly the diminishing return the graph shows.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 513-513
Trap
\[ s = \tfrac{4}{3} \cdot \sqrt{32} \approx 1.33 \cdot 5.66 \approx 7.53 \]
Round each factor to two decimal places, then multiply
Why: Two decimals seems plenty of precision.
Here it happens to come out close, but the exact answer is sixteen root two over three and rounding early throws that away. In a longer calculation the small errors accumulate and the final digit stops being trustworthy.
\[ s = \tfrac{16\sqrt{2}}{3} \approx 7.5 \text{ knots} \]
Keep the radical exact until the last line, then round once
Why: The exact form is the answer; the decimal is its interpretation.
The exact form can also be checked by squaring, which no rounded value can survive.
Sorting
Which form does each question want?
Sort into buckets
Sort each request by the form of answer it asks for.
Example 4 asks for both in one question, which is common. Giving only the decimal loses the exactness and giving only the radical does not answer how fast the boat goes.
Prediction
From 32 feet to 64 feet.
Predict first
What happens to the maximum speed?
Correct: It grows by a factor of about 1.41, not by a factor of 2.
\[ 32 \text{ ft}: \; 7.5 \text{ knots} \qquad 64 \text{ ft}: \; 10.7 \text{ knots} \]
Why: The speed is the square root of a constant times the length, so doubling the length multiplies the speed by the root of two. A boat twice as long is only about forty per cent faster, which is why hull length alone is a costly way to buy speed and why the graph flattens as it goes right.
Socratic
A sailor wants a number of knots.
Discussion prompt
Say why the exact radical is worth reporting alongside the rounded speed. Then say which one you would put on a specification sheet.
Hint: Ask what happens if the answer is used again.
Answer:
The radical is exact, so it can be checked by squaring and it can be used in further calculations without accumulating error. The rounded value cannot do either: squaring 7.5 gives 56.25 rather than the exact five hundred and twelve ninths, and feeding it into another step compounds the discrepancy.
A specification sheet would carry the rounded value with its units, because that is what a person acts on, along with the precision it was rounded to. The exact form belongs in the working, where anyone rechecking the design can see that no precision was lost before the final line.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Product property | Quotient property | |
|---|---|---|
| Statement | the root of ab is root a times root b | the root of a over b is root a over root b |
| Conditions | a and b both at least zero | a at least zero, b greater than zero |
| Used for | removing perfect square factors | removing a fraction from the radicand |
The stricter condition in the second column is there because a denominator cannot be nought. Everything else about the two rules is the same idea applied to a different operation.
Pattern
To put any radical expression into simplest form, work down this list.
Step three is the one most often skipped, and it is what separates two root eighteen from six root two — the same value, but only one of them in simplest form.
OpenStax Elementary Algebra 2e, §9.2 Simplify Square Roots §9.2
Check
Largest square factor.
Check your understanding
Write the root of 180 in simplest form.
Answer: A
Why: A hundred and eighty is thirty-six times five, and the root of thirty-six is six. Five is prime, so nothing more can be removed.
Check
Clear the denominator.
Check your understanding
Write 1 over the root of 7 in simplest form.
Answer: A
Why: Multiplying top and bottom by the root of seven makes the denominator seven and moves the radical upstairs, leaving about 0.378 as before.
Check
Reduce, then split.
Check your understanding
Write the root of 75 over 12 in simplest form.
Answer: A
Why: Seventy-five over twelve reduces to twenty-five over four, and both parts are perfect squares, so the radical disappears entirely.
Real world
This is the sailboat question from the lesson opener. The maximum speed s in knots of certain boats is modelled by s squared equals sixteen ninths times x, where x is the water line length in feet.
Discussion prompt
Express the maximum speed of a boat with a 32 foot water line exactly, then find it to the nearest tenth. Do the same for a 50 foot water line, and say what the comparison tells a designer.
Hint: Simplify the radical before rounding.
Answer:
\[ s = \sqrt{\tfrac{16}{9}(32)} = \tfrac{4}{3} \cdot 4\sqrt{2} = \tfrac{16\sqrt{2}}{3} \approx 7.5 \]
For fifty feet the same work gives twenty root two over three, which is about 9.4 knots.
Adding eighteen feet of water line — more than half again — bought only about 1.9 knots, a gain of about a quarter. Because speed depends on the square root of the length, each extra foot returns less than the one before, and a designer chasing speed eventually has to change something other than length. That is a genuine constraint that the square root in the model expresses precisely.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Which of these is in simplest form?
Correct: 6 root 2.
\[ 2\sqrt{18} = 2 \cdot 3\sqrt{2} = 6\sqrt{2} \]
Why: Two is prime, so no perfect square factor remains, there is no fraction under the radical and no radical in a denominator — all three conditions hold. Two root eighteen has the same value, about 8.49, but eighteen still contains a nine, so it fails the first condition. One over the root of two fails the third and becomes the root of two over two. The last option has a fraction under the radical and fails the second, becoming the root of five over four. Only one of the four is finished, and checking all three conditions rather than one is what settles it.
Explain it
They stopped at two times the root of eighteen and think they are finished.
Discussion prompt
In no more than four sentences, explain why that is not simplest form and how to spot the problem. Then give them a test they can run on any answer.
Hint: Look inside the radicand for a square.
Answer:
A usable answer: eighteen is nine times two, and nine is a perfect square, so a three is still trapped under the radical. Pulling it out gives two times three root two, which is six root two — the same number, properly finished.
The test is to factor the radicand into primes and look for any prime that appears twice. A repeated prime means a square is hiding and can come out; if no prime repeats, the expression is done.
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: Spotting factors is fixed by learning the squares up to twenty-five by sight. Knowing when to stop is fixed by factoring the radicand into primes and looking for repeats. Fractions under radicals are fixed by reducing before splitting. Rationalising is fixed by remembering the multiplier is always the offending radical over itself. 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 the three conditions for simplest form as a checklist with a box beside each, and write one expression that fails each condition next to the box it fails. Underneath, state the product and quotient properties with their conditions on a and b, and beside the product property write the counterexample showing that a root does not split over addition. In the middle, simplify the root of forty-eight twice, once using four as the factor and once using sixteen, writing both columns side by side so the extra steps in the first are visible. Beneath that, simplify a radical of a fraction that reduces to a ratio of perfect squares and one that does not, and rationalise two denominators showing the form of one you multiplied by each time. In the lower corner, list the perfect squares up to two hundred and twenty-five in a single row. Finally, in the margin, write the prime-factor test for deciding whether a radicand is finished.
Your two columns for the root of forty-eight should end on the same line. If they do not, one of them stopped before the radicand ran out of square factors.
Recap
Five things, and the last two are about the form of the answer rather than its value.
| If the question says | Your first move is |
|---|---|
| Simplify a radical of a whole number | Find its largest perfect square factor |
| The radicand is a fraction | Reduce it, then split with the quotient property |
| A radical sits in the denominator | Multiply by that radical over itself |
| Express in terms of radicals | Keep it exact; do not round |
| Find the value to the nearest tenth | Simplify first, then round once |
Lesson 9.4 leaves radicals behind and turns to graphs. Quadratic functions produce parabolas, and their shape explains why the equations of the last two lessons have two solutions, one, or none.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 9 Quadratic Equations and Functions — Lesson 9.3 Simplifying Radicals §9.3, pp. 511-516 — everything on these slides traces back here
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