Use square root and cube root symbols to solve x² = p and x³ = p and evaluate roots of perfect squares and cubes

Learning goal: Use square root and cube root symbols to represent solutions to equations of the form x² = p and x³ = p where p is a positive rational number, evaluate square roots of small perfect squares and cube roots of small perfect cubes, and know that √2 is irrational.

Perfect squares and the square root symbol. A number is a perfect square when it is the square of an integer: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121 and 144 are the squares of 1 through 12. The symbol √p, read "the square root of p", stands for the non-negative number whose square is p. So √49 = 7, √144 = 12 and √0.25 = 0.5, because 0.5² = 0.25. Evaluating a square root of a perfect square means recognizing which integer was squared. The number 0.25 is not a perfect square under this definition, since it is not the square of an integer, but its square root is still rational: a positive rational number has a rational square root exactly when, written as a fraction in lowest terms, its numerator and denominator are both perfect squares. So 0.25 = 1/4 gives √0.25 = 1/2 = 0.5, 2.25 = 9/4 gives √2.25 = 3/2 = 1.5, and √(9/16) = 3/4 because (3/4)² = 9/16.

Solving x² = p. The equation x² = 49 asks for every number whose square is 49. There are two: 7 and −7, because (−7)² = 49 as well. The solutions are written x = ±√49 = ±7. This is a point where the symbol and the equation differ, and the difference matters: √49 is the single number 7, while the equation x² = 49 has two solutions. For a positive rational p whose square root is not rational, which for p in lowest terms means that the numerator and denominator are not both perfect squares (so √(1/4) = 1/2 is rational, but √2 and √(3/5) are not), the solutions still exist but cannot be written as a fraction, and the root symbol is exactly how they are represented: x² = 2 has solutions x = ±√2, and x² = 3/5 has solutions x = ±√(3/5). Writing ±√2 is a complete and exact answer; a decimal such as 1.41 is only an approximation. Because the square of any real number is non-negative, x² = p has no real solutions when p is negative, and x² = 0 has the single solution 0.

Perfect cubes and the cube root symbol. A perfect cube is the cube of an integer: 1, 8, 27, 64, 125, 216, 343, 512, 729 and 1000 are the cubes of 1 through 10. The symbol ∛p, the cube root of p, is the number whose cube is p. So ∛64 = 4, ∛125 = 5 and ∛(8/27) = 2/3 because (2/3)³ = 8/27. Also ∛0.001 = 0.1 because 0.1³ = 0.001.

Solving x³ = p. Cubing does not lose the sign the way squaring does: 4³ = 64 but (−4)³ = −64. So for a positive rational p the equation x³ = p has exactly one real solution, x = ∛p, and no ± is needed. The equation x³ = 64 has the single solution x = 4, and x³ = 5 has the single solution x = ∛5. A negative right-hand side also gives one solution, x³ = −8 has x = −2, because a negative number does have a real cube root, unlike a real square root. This lesson focuses on positive p, as the goal states, but the sign behavior explains why square and cube root equations are treated differently.

Approximating a root that is not a perfect square or cube. When p is not a perfect square, √p is still a specific number that can be trapped between rationals. For x² = 12, the solutions are ±√12, and since 3² = 9 < 12 < 16 = 4², √12 is between 3 and 4; since 3.4² = 11.56 < 12 < 12.25 = 3.5², it is between 3.4 and 3.5. Refining, 3.46² = 11.9716 < 12 < 12.0409 = 3.47², and the midpoint test 3.465² = 12.006225 > 12 puts √12 below 3.465. Give the exact answer ±√12 first, then the approximation ±3.46 to the nearest hundredth if the problem asks for a decimal, noting that 3.46² = 11.9716, not exactly 12.

Why √2 is irrational — a worked argument inside the lesson. The number √2 is the side length of a square whose area is 2, and the diagram in this lesson shows it as the diagonal of a unit square: by the Pythagorean relationship, the diagonal squared is 1² + 1² = 2. Suppose, to test the idea, that √2 could be written as a fraction a/b of positive integers in lowest terms, so that a and b share no common factor larger than 1. Squaring both sides gives a²/b² = 2, so a² = 2b². The right side is even, so a² is even, and an odd number squared is always odd, so a itself must be even. Write a = 2k for some integer k. Then (2k)² = 4k² = 2b², so b² = 2k², which makes b² even and therefore b even. But now both a and b are even, sharing the factor 2, which contradicts the assumption that the fraction was in lowest terms. Since assuming √2 is a fraction leads to an impossibility, √2 is not a fraction: it is irrational. A similar style of argument, tracking a prime factor of p rather than even and odd, shows that √p is irrational whenever the whole number p is not a perfect square, and ∛p is irrational whenever the whole number p is not a perfect cube. The prime to track is one that appears in p an odd number of times for a square root, or a number of times that is not a multiple of 3 for a cube root: for √2 it is 2, and for √12, since 12 = 2 × 2 × 3, it is 3, not 2. This is the reason the root symbols are needed as exact notation, and it links this lesson to the classification of real numbers as rational or irrational: √49 = 7 and ∛8 = 2 are rational, while √2 and ∛2 are irrational, with decimal expansions that are non-repeating and non-terminating.

Common misconceptions. First, "√16 = ±4" is wrong as written: √16 is 4, and the equation x² = 16 is what has two solutions, ±4. Second, "√(a + b) = √a + √b" is false: √(9 + 16) = √25 = 5, while √9 + √16 = 3 + 4 = 7. The root of a product of non-negative numbers does split, √(9 × 16) = √9 × √16 = 3 × 4 = 12 = √144, and so does the root of a quotient with a non-negative numerator and a positive denominator, √(9/16) = 3/4, but splitting the root of a sum is not a valid rule (it gives the right value only in the trivial case where one of the two numbers is 0). Third, halving is not taking a square root: √100 is 10, not 50; a square root asks which number multiplied by itself gives 100. Fourth, √2 ≈ 1.414 is not equal to 1.414; the exact value has no finite decimal. Fifth, a cube root is not "a third of the number": ∛27 = 3 because 3 × 3 × 3 = 27, not because 27 ÷ 3 = 9.

Checking solutions. Every solution can be substituted back. For x² = 1.44, the candidates are ±1.2, and 1.2² = 1.44 and (−1.2)² = 1.44 both check. For x³ = 8/125, the candidate 2/5 gives (2/5)³ = 8/125, which checks, and −2/5 gives −8/125, which does not. Substitution catches an extra sign and any arithmetic slip, because a wrong candidate fails the check. It cannot catch a solution that was never written down, because an omitted candidate is never tested: answering only x = 1.2 for x² = 1.44 passes substitution. To be sure none is missing, use the form of the equation: for a positive p, x² = p always has exactly two solutions, ±√p, and x³ = p has exactly one, ∛p.

Worked square root equation with a perfect square: Solve x² = 81. The solutions are x = ±√81 = ±9, because 9² = 81 and (−9)² = 81. Note the distinction: √81 alone is 9, but the equation has two solutions. Both check by substitution.

Worked cube root equation with a fraction: Solve x³ = 27/64. Both 27 and 64 are perfect cubes, 3³ and 4³, so x = ∛(27/64) = 3/4. There is only one real solution because the cube of −3/4 is −27/64, not 27/64. Check: (3/4)³ = 27/64.

Worked equation whose solution is irrational: Solve x² = 20 and give the solutions exactly and approximately. Exactly, x = ±√20. Since 20 is not a perfect square, √20 is irrational and the root symbol is the exact form. To approximate, 4² = 16 < 20 < 25 = 5², so √20 is between 4 and 5; 4.4² = 19.36 < 20 < 20.25 = 4.5², so it is between 4.4 and 4.5; 4.47² = 19.9809 < 20 < 20.0704 = 4.48², and the midpoint test 4.475² = 20.025625 > 20 puts √20 below 4.475, so to the nearest hundredth √20 ≈ 4.47. The solutions are approximately ±4.47, and the exact answer stays ±√20.

Worked geometry with a cube: A cube-shaped box has volume 216 cubic centimeters. Its edge length e satisfies e³ = 216, so e = ∛216 = 6 cm, because 6 × 6 × 6 = 216. The negative cube root does not exist for a positive number, and a negative length would be meaningless anyway, so 6 cm is the only answer. A square with area 216 cm² would instead have side √216 cm, which is irrational and lies between 14 and 15 because 14² = 196 and 15² = 225.

Root reference record: Build two tables from your own calculations: the squares of 1 through 15 and the cubes of 1 through 10. Beside each table, write five equations of the form x² = p or x³ = p whose solutions can be read from the table, and record the solution set for each, using ± only where it belongs. Then draw a unit square with its diagonal, label the sides 1 and the diagonal √2, and write the two-line justification for why the diagonal squared is 2.

Independent performance: Write and solve eight equations: three of the form x² = p with p a perfect square, a fraction of perfect squares and a decimal such as 0.49; three of the form x³ = p with similar variety; and two where p is a positive rational number that is not a perfect square or cube, giving exact root-symbol answers and a two-decimal approximation trapped between squared or cubed candidates. Then write the √2 irrationality argument in your own words in no more than eight sentences. Success criteria: every solution is checked by substitution, square equations show two solutions and cube equations show one, exact and approximate answers are labeled with = and ≈ respectively, and the argument states clearly where the contradiction appears.