Radicals, Rational Exponents, and Complex Numbers

Square roots, fractional powers, and complex numbers extend familiar exponent rules and quadratic solving. This lesson assumes comfort with factoring and the quadratic formula from Algebra. The central habit is to state the number system and domain before manipulating an expression: a step that is valid for real numbers may not be valid for every complex number.

Radicals and Their Real-Number Domains

The symbol √a means the principal square root: the one nonnegative real number whose square is a. Thus √49 = 7, although both 7 and −7 solve x2 = 49. Over the real numbers, √a is defined only when a ≥ 0. An even-index radical such as 4√a has the same restriction; an odd-index radical such as 3√a is defined for every real a.

To simplify a radical, factor its radicand into a perfect-square factor and a remaining factor. For real variables, do not forget that √(u2) = |u|, not always u.

Worked example: simplify with a variable

√(72x⁵), with x ≥ 0
= √(36x⁴ · 2x)
= 6x²√(2x)

The condition x ≥ 0 is needed for the original real square root.

If the condition had instead been x ≤ 0, then √(x2) would be |x| = −x. Absolute value records the nonnegative output of a square root.

Worked example: add only like radicals

3√12 − 2√27 + √3
= 3(2√3) − 2(3√3) + √3
= 6√3 − 6√3 + √3
= √3

Radicals can be added only after their simplified radical parts match, just as 3x and 5x are like terms but 3x and 5y are not.

Rational Exponents

For a positive real base a and positive integers m and n,

am/n = n√(am) = (n√a)m.

The denominator n is the root index; the numerator m is the power. A negative exponent takes a reciprocal: a−m/n = 1/am/n, so a cannot be zero in that expression.

For real-valued expressions, an even root requires a nonnegative radicand. For example, x2/3 is real for every real x, because the cube root comes first; x3/2 requires x ≥ 0. For a negative base and a rational exponent m/n, first reduce the fraction. A real value is defined when the reduced denominator n is odd, using the real odd root and then the power: (−8)2/3 = (∛(−8))2 = 4. If the reduced denominator is even, there is no real value. Do not choose an unreduced root index mechanically: although 2/3 = 4/6, interpreting (−8)4/6 as a sixth root would incorrectly make the same exponent undefined. Outside this reduced-fraction, odd-root convention, fractional powers of negative bases can be ambiguous, so radical notation is clearest.

Worked example: evaluate and rewrite

16^(3/4) · 27^(−2/3)
= (⁴√16)³ · 1/(∛27)²
= 2³ · 1/3²
= 8/9

x^(5/2) = x²√x, with x ≥ 0

Solving Radical Equations Safely

First state restrictions from every even radical and every denominator. Then isolate a radical, raise both sides to the needed power, solve, and check every candidate in the original equation. Squaring is not reversible: if a = b, then a2 = b2; but equal squares can also come from opposites.

Worked example: an extraneous root

Solve √(x + 5) = x − 1 over the reals.

Restrictions: x + 5 ≥ 0, and the right side must be ≥ 0.
So x ≥ 1.

Square: x + 5 = (x − 1)²
         x + 5 = x² − 2x + 1
         0 = x² − 3x − 4
         0 = (x − 4)(x + 1)

Candidates: x = 4, −1
Check: √(4 + 5) = 3 = 4 − 1, so x = 4 works.
       √(−1 + 5) = 2, but −1 − 1 = −2, so x = −1 is extraneous.

Solution: x = 4

Worked example: a radical with a denominator restriction

Solve √(x − 1)/(x − 3) = 1.

Original domain: x ≥ 1 and x ≠ 3.

√(x − 1) = x − 3
The right side must be nonnegative, so x ≥ 3; together, x > 3.

x − 1 = (x − 3)²
0 = x² − 7x + 10
0 = (x − 5)(x − 2)

Only x = 5 meets x > 3.
Check: √4/(5 − 3) = 2/2 = 1.
Solution: x = 5

The Imaginary Unit and Complex Numbers

No real number squares to −1. To solve equations that require such a number, define the imaginary unit i by i2 = −1. A complex number has the form a + bi, where a and b are real. Its real part is a; its imaginary part is b. Every real number is complex with imaginary part 0.

In complex arithmetic, simplify powers of i using the cycle i, −1, −i, 1, which repeats every four powers. The symbol √(−9) is not a real number; in the complex number system its principal value is 3i.

Worked example: multiply and divide complex numbers

(3 − 2i)(4 + i)
= 12 + 3i − 8i − 2i²
= 12 − 5i + 2
= 14 − 5i

(5 + 3i)/(1 − 2i)
= (5 + 3i)(1 + 2i) / [(1 − 2i)(1 + 2i)]
= (5 + 10i + 3i + 6i²) / (1 − (2i)²)
= (−1 + 13i)/5
= −1/5 + 13/5 i

The factor 1 + 2i is the conjugate of 1 − 2i. In general, (a + bi)(a − bi) = a2 + b2, a nonzero real number unless both parts are zero. Thus a complex denominator may not be 0 + 0i.

Quadratic Roots, Including Nonreal Roots

For ax2 + bx + c = 0 with a ≠ 0, the quadratic formula is x = (−b ± √(b2 − 4ac))/(2a). The discriminant D = b2 − 4ac predicts the roots: D > 0 gives two real roots, D = 0 one repeated real root, and D < 0 two complex conjugate roots.

Worked example: solve a quadratic with negative discriminant

Solve 2x² + 4x + 5 = 0.

D = 4² − 4(2)(5) = 16 − 40 = −24
x = [−4 ± √(−24)]/4
  = [−4 ± 2i√6]/4
  = −1 ± (√6/2)i

The two roots are conjugates: −1 + (√6/2)i and −1 − (√6/2)i.

Common Mistakes to Avoid

  • Dropping absolute value: √(x2) = |x|. For example, when x = −3, the left side is 3, not −3.
  • Splitting a sum under a radical: √(a + b) is generally not √a + √b. For instance, √(9 + 16) = 5, not 7.
  • Keeping all squared-equation answers: candidates from squaring must be checked against the original equation.
  • Forgetting a denominator restriction: simplifying cannot restore a value that made the original denominator zero.
  • Treating i as a variable: replace i2 by −1; do not leave an answer with an i2 term.

Practice Set

  1. Simplify √(200x3) for x ≥ 0.
  2. Simplify 4√18 + √8 − 3√2.
  3. State the real domain of f(x) = √(5 − 2x)/(x + 1).
  4. Rewrite (x7)1/3 using a radical, and state its real domain.
  5. Evaluate 813/4 / 82/3.
  6. Solve √(2x + 3) = x.
  7. Solve √(x + 6) + 1 = x.
  8. Simplify i37 and i58.
  9. Compute (2 − 3i) + (−5 + 7i) and (2 − 3i)(−5 + 7i).
  10. Write (4 + i)/(2 − i) in a + bi form.
  11. Solve x2 − 6x + 13 = 0.
  12. For 3x2 + 2x + k = 0, find the value of k that gives one repeated real root.

Complete Answers

  1. √(200x3) = √(100x2 · 2x) = 10x√(2x), using x ≥ 0.
  2. 4√18 + √8 − 3√2 = 4(3√2) + 2√2 − 3√2 = 11√2.
  3. Require 5 − 2x ≥ 0 and x + 1 ≠ 0. Thus x ≤ 5/2, excluding −1: (−∞, −1) ∪ (−1, 5/2].
  4. (x7)1/3 = ∛(x7) = x2∛x. Cube roots accept all real inputs, so the domain is all real numbers.
  5. 813/4 / 82/3 = (⁴√81)3/(∛8)2 = 33/22 = 27/4.
  6. The right side requires x ≥ 0. Squaring gives 2x + 3 = x2, or (x − 3)(x + 1) = 0. Check: x = 3 works; x = −1 does not. Solution: 3.
  7. Isolate: √(x + 6) = x − 1, so x ≥ 1. Squaring gives x + 6 = x2 − 2x + 1, or (x − 3)(x + 2) = 0. Only 3 satisfies the original equation.
  8. Since powers repeat every four, i37 = i and i58 = i2 = −1.
  9. The sum is −3 + 4i. The product is −10 + 14i + 15i − 21i2 = 11 + 29i.
  10. Multiply by 2 + i: (4 + i)(2 + i)/[(2 − i)(2 + i)] = (8 + 6i + i2)/5 = 7/5 + 6/5i.
  11. D = (−6)2 − 4(1)(13) = −16. Therefore x = [6 ± 4i]/2 = 3 ± 2i.
  12. A repeated real root requires discriminant 0: 22 − 4(3)k = 0. Hence 4 − 12k = 0 and k = 1/3.