Simplifying a square root
A square floor covers 245 square meters. Its side is the square root of 245, but can that number be written exactly and still show how big it is?
Where you are 3 in Simplifying radicals
When you reach for it
Maximilian is marking out a square dance floor with an area of 245 square meters. The side of a square is the number whose square is its area, so the side is \displaystyle \sqrt{245} meters. That is exact, but it hides the size: is it near 15, near 16, a whole number? A calculator gives 15.6524\ldots, which shows the size and is no longer exact.
There is a third way to write it. Since 245 = 49 \cdot 5, the floor can be cut into 49 small squares of area 5, seven along each side. Each small square has side \displaystyle \sqrt{5}, so the side of the floor is seven of those:
\displaystyle 7\sqrt{5}, read “seven root five”, means \displaystyle 7 \cdot \sqrt{5}. It is exactly the same number as \displaystyle \sqrt{245}, written so its size shows: \displaystyle \sqrt{5} \approx 2.236, so the side is about 7 \times 2.236 \approx 15.65 meters.
Reach for this rewrite whenever a square root is not a whole number and you want to keep it exact: the side of a square from its area, a length from Pythagoras’ theorem, the answers that come out of the square root property and the quadratic formula. Simplified roots can be compared at a glance, estimated by hand, and recognized as multiples of the same root, which is what later lets them be added.
The words
| Word | What it names | In \displaystyle \sqrt{245} | In \displaystyle \sqrt[3]{250} |
|---|---|---|---|
| radical sign | the symbol itself | \displaystyle \sqrt{\ } | \displaystyle \sqrt[3]{\ } |
| radicand | what sits under the sign | 245 | 250 |
| index | the small number in the crook of the sign, saying which root | 2, not written | 3 |
| radical | the whole expression | \displaystyle \sqrt{245} | \displaystyle \sqrt[3]{250} |
When no index is written it is 2, a square root. For a \ge 0, \displaystyle \sqrt{a} is the non-negative number whose square is a (square and cube roots). An index of 3 asks for a cube root. This lesson is about index 2, with a short note on index 3 at the end of the procedure.
The procedure
The product rule
Everything here rests on one fact. For any numbers a \ge 0 and b \ge 0,
Why it is true. By definition, \displaystyle \sqrt{ab} is the one non-negative number whose square is ab. So it is enough to show that \displaystyle \sqrt{a}\,\sqrt{b} is non-negative and squares to ab.
It is non-negative because \displaystyle \sqrt{a} and \displaystyle \sqrt{b} both are, and a product of non-negative numbers is non-negative. Its square is
where the middle step reorders the factors (the commutative property) and regroups them (the associative property). Only one non-negative number has a given square: if p \ge 0, q \ge 0 and p^2 = q^2, then (p - q)(p + q) = 0, so p = q or p + q = 0, and the second happens only when p = q = 0. So the two sides are the same number.
Read left to right, the rule splits a root. When one factor is a perfect square k^2 with k \ge 0, its root is simply k:
That is the whole skill: find a perfect-square factor of the radicand and bring its root outside.
Simplest form
A square root is in simplest form when its radicand has no perfect-square factor other than 1. So \displaystyle \sqrt{5}, \displaystyle \sqrt{15} and \displaystyle \sqrt{42} are in simplest form, and \displaystyle \sqrt{180} is not, because 36 is a factor of 180. The whole number goes in front: \displaystyle 6\sqrt{5}, not \displaystyle \sqrt{5} \cdot 6. (Roots in fractions bring one more condition, no root left in a denominator, which belongs to rationalizing a denominator.)
Taking out the largest square
To simplify \displaystyle \sqrt{180}, run through the perfect squares below 180, which are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, and find the largest that divides 180. It is 36, since 180 = 36 \cdot 5 and none of the larger ones divides 180. So
The radicand left, 5, has no square factor, so this is simplest form.
Or in steps
You do not have to see the largest square at once. Any square factor you notice can come out, and you carry on until none is left. 180 is a multiple of 4:
45 still has the square factor 9, so
Each step is the product rule again, and the route ends where the first one did.
Prime factorization finds the pairs
For a large radicand, a prime factorization does the searching:
Each pair of equal primes is a perfect square and leaves the root as one copy, since \displaystyle \sqrt{2^2} = 2 and \displaystyle \sqrt{3^2} = 3. A prime with no partner stays inside. Outside: 2 \cdot 3 = 6. Inside: 5. So \displaystyle \sqrt{180} = 6\sqrt{5}.
This also shows why every route gives the same answer. A whole number has only one prime factorization, so which primes appear an odd number of times is fixed. However the squares are taken out, one copy of each of those primes is what stays under the root, and one prime from every pair is what comes out. So “the” simplest form is a fair name: there is only one.
Variables
In this lesson every variable under a square root stands for a non-negative number. The assumption is needed: \displaystyle \sqrt{x^2} = x is true when x \ge 0, but for x = -3, \displaystyle \sqrt{(-3)^2} = \sqrt{9} = 3, not -3.
With it, an even power comes out at half its exponent, \displaystyle \sqrt{x^{2k}} = x^k, because \left(x^k\right)^2 = x^{2k} by the power of a power rule and x^k \ge 0. An odd power gives up one factor first: x^7 = x^6 \cdot x, so \displaystyle \sqrt{x^7} = x^3\sqrt{x}. It is the pairing idea again, with x^7 as seven equal factors.
Checking with a decimal
A calculator check is quick: \displaystyle \sqrt{180} \approx 13.416 and \displaystyle 6\sqrt{5} \approx 6 \times 2.236 = 13.416. A check by hand works too. 13^2 = 169 and 14^2 = 196, so \displaystyle \sqrt{180} lies between 13 and 14 (estimating a square root), and \displaystyle 6\sqrt{5}, with \displaystyle \sqrt{5} between 2.2 and 2.3, lies between 13.2 and 13.8. A decimal check catches a wrong factor. It cannot tell you the form is simplest; only the square-factor test does that.
Cube roots, briefly
The same move works for cube roots, with perfect cubes (8, 27, 64, 125, \ldots) in place of squares. The product rule \displaystyle \sqrt[3]{ab} = \sqrt[3]{a}\,\sqrt[3]{b} has the same proof with cubes, and it holds for negative numbers too, since every real number has exactly one real cube root.
In a prime factorization, each group of three equal primes comes out as one. Any index is handled in higher roots.
Worked examples
Taking out the largest square factor
Simplify (a) \displaystyle \sqrt{368} \qquad (b) \displaystyle \sqrt{135} \qquad (c) \displaystyle \sqrt{243}
Solution(a) The perfect squares that divide 368 are 4 and 16. The larger is 16, and 368 = 16 \cdot 23:
Taking 4 first works too, just in two steps: \displaystyle \sqrt{368} = 2\sqrt{92} = 2 \cdot 2\sqrt{23} = 4\sqrt{23}. As a check, 4 \times 4.796 = 19.184, and 19.184^2 \approx 368.0.
(b) 135 = 9 \cdot 15, and no larger square (25, 49, 81, 121) divides 135:
15 = 3 \cdot 5 has one 3 and one 5, no pair, so it has no square factor and \displaystyle 3\sqrt{15} is simplest.
(c) 243 = 81 \cdot 3, and 81 is the largest square below 243 that divides it:
The same root by steps and by primes
Simplify \displaystyle \sqrt{864} in steps, then again with a prime factorization.
SolutionIn steps. 864 is a multiple of 4, and so is the quotient, and then 9 appears:
6 = 2 \cdot 3 has no square factor, so this is simplest.
By primes. 864 = 2^5 \cdot 3^3, that is
There are two pairs of 2s and one pair of 3s, so 2 \cdot 2 \cdot 3 = 12 comes out. The unpaired 2 and 3 stay in, as 6. Again \displaystyle \sqrt{864} = 12\sqrt{6}. The primes also say that 12^2 = 144 was the largest square factor all along.
As a check, \displaystyle 12\sqrt{6} \approx 12 \times 2.449 = 29.39, and \displaystyle \sqrt{864} \approx 29.39.
The diagonal of an origami sheet
Loveday folds a square origami sheet of side 18 cm along a diagonal. How long is the fold? Give it exactly, in simplest form, and to the nearest hundredth of a centimeter.
Two sides of the square meet at a right angle, so the diagonal is the hypotenuse of a right triangle with legs 18 and 18. By Pythagoras’ theorem,
and \displaystyle d = \sqrt{648}, the positive root, since d is a length. Now 648 = 2^3 \cdot 3^4: one pair of 2s and two pairs of 3s come out as 2 \cdot 3 \cdot 3 = 18, and one 2 stays in. So the fold is \displaystyle 18\sqrt{2} cm, about 18 \times 1.4142 \approx 25.46 cm.
There was no need to multiply out. d^2 = 18^2 + 18^2 = 2 \cdot 18^2, so \displaystyle d = \sqrt{18^2 \cdot 2} = 18\sqrt{2} straight away. The same line shows that any square of side s has diagonal \displaystyle s\sqrt{2}, the fact behind the 45-45-90 triangle.
Variables under the root
Simplify, taking every variable to be non-negative.
(a) \displaystyle \sqrt{162x^3y^6} \qquad (b) \displaystyle \sqrt{240a^9b}
SolutionSplit each factor into its largest square part and what is left.
(a) 162 = 81 \cdot 2, x^3 = x^2 \cdot x, and y^6 = \left(y^3\right)^2 is already a square.
The first root is 9xy^3 because \left(9xy^3\right)^2 = 81x^2y^6 and 9xy^3 \ge 0 when x and y are.
(b) 240 = 16 \cdot 15, a^9 = a^8 \cdot a, and b has no square part.
What is left, 15ab, has no square factor: 15 = 3 \cdot 5, and a and b appear once each.
Comparing, and spotting multiples of one root
(a) Without a calculator, put \displaystyle \sqrt{47}, \displaystyle 3\sqrt{5} and \displaystyle 2\sqrt{11} in increasing order.
(b) Which of \displaystyle \sqrt{175}, \displaystyle \sqrt{294}, \displaystyle \sqrt{343} and \displaystyle \sqrt{448} are whole-number multiples of \displaystyle \sqrt{7}?
Solution(a) Read the product rule right to left to move each number back under the root: for k \ge 0, \displaystyle k\sqrt{m} = \sqrt{k^2}\,\sqrt{m} = \sqrt{k^2 m}. So \displaystyle 2\sqrt{11} = \sqrt{44} and \displaystyle 3\sqrt{5} = \sqrt{45}. For non-negative numbers, a larger radicand gives a larger square root, because squaring keeps the order of non-negative numbers. So \displaystyle 2\sqrt{11} < 3\sqrt{5} < \sqrt{47}.
(b) Simplify each one:
Three are: \displaystyle \sqrt{175}, \displaystyle \sqrt{343} and \displaystyle \sqrt{448}. The fourth is not, since \displaystyle 7\sqrt{6} = k\sqrt{7} would give, on squaring, 294 = 7k^2, so k^2 = 42, and no whole number squares to 42. None of this was visible before simplifying. Adding such roots, as in \displaystyle 5\sqrt{7} + 8\sqrt{7}, is taken up in operations with radicals.
A root inside a larger expression
Simplify \displaystyle \dfrac{10 + \sqrt{300}}{5}, and check the result with decimals.
SolutionFirst the root: \displaystyle \sqrt{300} = \sqrt{100 \cdot 3} = 10\sqrt{3}. So
Dividing a sum by 5 divides each term by 5 (the distributive property). The two terms 2 and \displaystyle 2\sqrt{3} cannot be merged into one number: \displaystyle \sqrt{3} is irrational, so the sum stays as it is, in the same way 2 + 2x does.
Check: \displaystyle \sqrt{300} \approx 17.321, so the original is about \displaystyle \frac{27.321}{5} \approx 5.464, and 2 + 2 \times 1.732 = 5.464. Answers of exactly this shape come out of the quadratic formula, and simplifying the root first is what lets the fraction cancel.
Boundaries
- The product rule needs a \ge 0 and b \ge 0. (-4)(-9) = 36, so \displaystyle \sqrt{(-4)(-9)} = \sqrt{36} = 6. But no real number squares to a negative number, so \displaystyle \sqrt{-4} and \displaystyle \sqrt{-9} are not real numbers, and in the reals \displaystyle \sqrt{-4}\,\sqrt{-9} has no value at all. A root splits only into non-negative factors.
- Square roots do not split over sums or differences. \displaystyle \sqrt{9 + 16} = \sqrt{25} = 5, while \displaystyle \sqrt{9} + \sqrt{16} = 3 + 4 = 7. And \displaystyle \sqrt{25 - 9} = \sqrt{16} = 4, while \displaystyle \sqrt{25} - \sqrt{9} = 2. The product rule is about factors only, which is why \displaystyle \sqrt{x^2 + y^2} does not simplify.
- \displaystyle \sqrt{x^2} = x needs x \ge 0. That is why the non-negative assumption on variables is stated rather than left implicit.
- Simplest form is about square factors only. \displaystyle \sqrt{30} = \sqrt{2 \cdot 3 \cdot 5} is already simplest. Writing it as \displaystyle \sqrt{2}\,\sqrt{15} is true but not simpler.
- For a cube root the test is cube factors. \displaystyle \sqrt[3]{36} has the square factor 4 but no cube factor other than 1, so as a cube root it is already simplest.
Where this goes next
With roots in simplest form, operations with radicals can multiply and divide them and add the ones that are multiples of the same root, such as \displaystyle 5\sqrt{7} and \displaystyle 8\sqrt{7}. A root left in a denominator is cleared by rationalizing the denominator. Simplified roots are also the expected form for answers from the square root property and the quadratic formula, and for lengths found with Pythagoras’ theorem. Equations with the variable under the root are solved in solving radical equations. Further on, \displaystyle \sqrt{a} is written as the power a^{1/2} (rational exponents), and the product rule turns out to be a law of exponents in disguise.
Exercises
Words and simplest form
- Name the radicand and the index of (a) \displaystyle \sqrt{91} and (b) \displaystyle \sqrt[3]{x + 7}.
- Which of \displaystyle \sqrt{66}, \displaystyle \sqrt{99}, \displaystyle \sqrt{105} and \displaystyle \sqrt{275} are in simplest form? Simplify the others.
Simplify
- \displaystyle \sqrt{54}
- \displaystyle \sqrt{126}
- \displaystyle \sqrt{338}
- \displaystyle \sqrt{405}
- \displaystyle \sqrt{588}
Compare
- Without a calculator, put \displaystyle 5\sqrt{2}, \displaystyle 4\sqrt{3} and \displaystyle 2\sqrt{13} in increasing order.
With a prime factorization
- \displaystyle \sqrt{1512}
- \displaystyle \sqrt{2028}
Variables
Take every variable to be non-negative.
- \displaystyle \sqrt{w^9}
- \displaystyle \sqrt{76p^4q^3}
- \displaystyle \sqrt{392x^7y^{10}}
Cube roots
- Simplify (a) \displaystyle \sqrt[3]{40} and (b) \displaystyle \sqrt[3]{135}.
Problems
- Lisette’s square dance floor has an area of 288 square meters. Find its side exactly, in simplest form, and to the nearest hundredth of a meter.
- A square origami sheet of side 21 cm is folded along a diagonal. Find the length of the fold exactly, in simplest form, and to the nearest hundredth of a centimeter.
Reasoning
- Compute \displaystyle \sqrt{36 + 64} and \displaystyle \sqrt{36} + \sqrt{64}. Then use the square of \displaystyle \sqrt{a} + \sqrt{b} to explain why there is no “sum rule” to go with the product rule.
- Meraud simplifies \displaystyle \sqrt{1350} by taking out 25 first, and Lucan by taking out 9 first. Carry out both routes, and explain with the prime factorization of 1350 why they had to agree.
Answers
- (a) Radicand 91, index 2 (not written). (b) Radicand x + 7, index 3.
- \displaystyle \sqrt{66} and \displaystyle \sqrt{105} are in simplest form, since 66 = 2 \cdot 3 \cdot 11 and 105 = 3 \cdot 5 \cdot 7 have no repeated prime. The others: \displaystyle \sqrt{99} = \sqrt{9 \cdot 11} = 3\sqrt{11} and \displaystyle \sqrt{275} = \sqrt{25 \cdot 11} = 5\sqrt{11}.
- \displaystyle 3\sqrt{6}, since 54 = 9 \cdot 6.
- \displaystyle 3\sqrt{14}, since 126 = 9 \cdot 14 and 14 = 2 \cdot 7 has no square factor.
- \displaystyle 13\sqrt{2}, since 338 = 169 \cdot 2.
- \displaystyle 9\sqrt{5}, since 405 = 81 \cdot 5.
- \displaystyle 14\sqrt{3}, since 588 = 196 \cdot 3.
- \displaystyle 4\sqrt{3} < 5\sqrt{2} < 2\sqrt{13}. Moving each number under the root gives \displaystyle \sqrt{48}, \displaystyle \sqrt{50} and \displaystyle \sqrt{52}.
- \displaystyle 6\sqrt{42}. 1512 = 2^3 \cdot 3^3 \cdot 7: one pair of 2s and one pair of 3s give 2 \cdot 3 = 6 outside, and 2 \cdot 3 \cdot 7 = 42 stays inside.
- \displaystyle 26\sqrt{3}. 2028 = 2^2 \cdot 3 \cdot 13^2: the pairs give 2 \cdot 13 = 26, and 3 stays.
- \displaystyle w^4\sqrt{w}, since w^9 = w^8 \cdot w and \displaystyle \sqrt{w^8} = w^4.
- \displaystyle 2p^2q\sqrt{19q}, since 76p^4q^3 = \left(4p^4q^2\right)(19q).
- \displaystyle 14x^3y^5\sqrt{2x}, since 392x^7y^{10} = \left(196x^6y^{10}\right)(2x).
- (a) \displaystyle 2\sqrt[3]{5}, since 40 = 8 \cdot 5. (b) \displaystyle 3\sqrt[3]{5}, since 135 = 27 \cdot 5.
- \displaystyle 12\sqrt{2} m, about 16.97 m. The side is \displaystyle \sqrt{288} = \sqrt{144 \cdot 2}.
- \displaystyle 21\sqrt{2} cm, about 29.70 cm. The fold is \displaystyle \sqrt{21^2 + 21^2} = \sqrt{2 \cdot 21^2}, which is \displaystyle \sqrt{882}.
- \displaystyle \sqrt{36 + 64} = \sqrt{100} = 10, while \displaystyle \sqrt{36} + \sqrt{64} = 6 + 8 = 14. The product rule holds because \displaystyle \sqrt{a}\,\sqrt{b} squares to exactly ab. But \displaystyle \left(\sqrt{a} + \sqrt{b}\right)^2 = a + 2\sqrt{a}\,\sqrt{b} + b, which is more than a + b by \displaystyle 2\sqrt{ab}. So \displaystyle \sqrt{a} + \sqrt{b} is the square root of a + b only when that extra term is 0, that is, when a = 0 or b = 0.
- Both reach \displaystyle 15\sqrt{6}. Meraud: \displaystyle \sqrt{1350} = 5\sqrt{54} = 5 \cdot 3\sqrt{6} = 15\sqrt{6}. Lucan: \displaystyle \sqrt{1350} = 3\sqrt{150} = 3 \cdot 5\sqrt{6} = 15\sqrt{6}. Since 1350 = 2 \cdot 3^3 \cdot 5^2, the unpaired primes are one 2 and one 3, so 6 must stay under the root on any route, and the pairs of 3s and 5s must give 3 \cdot 5 = 15 outside. The prime factorization is unique, so no route can end anywhere else.