Radicals & Rational Exponents

A cube root has an odd index (3), so a negative radicand is perfectly allowed.

• Ask: what number cubed equals −8? → \((-2)^3 = -8\) ✓
• Therefore \(\sqrt[3]{-8} = \mathbf{-2}\)

Key rule: Odd index → result keeps the same sign as the radicand.

The radical symbol always returns the principal (positive) root.

• Find a positive number raised to the 4th power equaling 256: \(4^4 = 256\) ✓
• So \(\sqrt[4]{256} = \mathbf{4}\)

Watch out: \(-4\) also satisfies \((-4)^4 = 256\), but the radical symbol only returns the positive principal root.

The trap: \(-27^{4/3}\) means \(-(27^{4/3})\) — the negative is applied last, not to the base.

• Cube root of 27: \(27^{1/3} = 3\)
• Raise to the 4th power: \(3^4 = 81\)
• Apply the outside negative: \(-(81) = \mathbf{-81}\)

If it were \((-27)^{4/3}\): \((\sqrt[3]{-27})^4 = (-3)^4 = +81\) — a completely different answer!

An even-index root of a negative number does not exist in real numbers.

• \(\sqrt[4]{-16}\) → even index (4), negative radicand → Not a real number ✗
• \(\sqrt[3]{-8} = -2\) → odd index (3) → fine ✓
• \(-\sqrt[4]{16} = -2\) → radicand is +16, the negative is outside → fine ✓
• \(\sqrt[5]{-32} = -2\) → odd index (5) → fine ✓

The parentheses mean −27 is the entire base, so the exponent applies to it.

• Take the cube root (index 3 is odd → negative base OK): \(\sqrt[3]{-27} = -3\)
• Raise to the 4th power (numerator): \((-3)^4 = 81\)
• Result: \(\mathbf{81}\)

One pair of parentheses changes everything: \(-27^{4/3} = -81\) but \((-27)^{4/3} = 81\)!

Same base (3) in a division → subtract the exponents.

• Subtract: \(\dfrac{2}{3} - \left(-\dfrac{1}{3}\right) = \dfrac{2}{3} + \dfrac{1}{3} = 1\)
• Result: \(3^1 = \mathbf{3}\)

Trap: Subtracting a negative flips to addition. Don't forget to flip the sign of \(-\tfrac{1}{3}\)!

Same base (\(y\)) being multiplied → add the exponents.

• Convert 2 to fourths: \(2 = \dfrac{8}{4}\)
• Add: \(\dfrac{8}{4} + \dfrac{3}{4} = \dfrac{11}{4}\)
• Result: \(y^{11/4}\)

Same base (\(x\)) multiplied → add the exponents.

• \(x^{15} \cdot x^{20} = x^{15+20} = \mathbf{x^{35}}\)

The most fundamental exponent multiplication rule.

Same base divided → subtract the denominator exponent from the numerator.

• \(x^{5/3} \div x^{2/3} = x^{5/3 - 2/3}\)
• \(= x^{3/3} = x^1 = \mathbf{x}\)

Trap: Adding \(\tfrac{5}{3} + \tfrac{2}{3} = \tfrac{7}{3}\) is wrong here. Division always means subtraction of exponents!

A negative exponent means flip to the denominator: \(x^{-n} = \dfrac{1}{x^n}\)

• Flip: \(x^{-3/4} = \dfrac{1}{x^{3/4}}\)
• Rewrite in radical form: \(x^{3/4} = \sqrt[4]{x^3}\)
• Final: \(\dfrac{1}{\sqrt[4]{x^3}}\)

Remember: A negative exponent does NOT make the value negative — it creates a reciprocal!

Simplify each radical first, then combine like terms.

• \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\)  →  \(8 \times 2\sqrt{2} = 16\sqrt{2}\)
• \(\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}\)  →  \(4 \times 3\sqrt{2} = 12\sqrt{2}\)
• Combine: \(16\sqrt{2} - 12\sqrt{2} = \mathbf{4\sqrt{2}}\)

You cannot add \(\sqrt{12}\) and \(\sqrt{27}\) directly — simplify each one first.

• \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\)  →  \(3 \times 2\sqrt{3} = 6\sqrt{3}\)
• \(\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}\)  →  \(2 \times 3\sqrt{3} = 6\sqrt{3}\)
• Combine like terms: \(6\sqrt{3} + 6\sqrt{3} = \mathbf{12\sqrt{3}}\)

Both radicals now share \(\sqrt{3}\), so they combine just like variable terms.

Break into a perfect-square factor and a leftover factor.

• \(\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}\)
• \(\sqrt{x^4} = x^{4/2} = x^2\)
• Combine: \(\sqrt{50x^4} = \mathbf{5x^2\sqrt{2}}\)

\(\sqrt{a+b} = \sqrt{a}+\sqrt{b}\) is always false.

• Counter-example: \(\sqrt{9+16} = \sqrt{25} = 5\)
• But \(\sqrt{9}+\sqrt{16} = 3+4 = 7 \neq 5\) ✗

Valid rules: \(\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}\) ✓   and   \(\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}\) ✓
Radicals distribute over multiplication and division — never over addition or subtraction.

When dividing radicals with the same index, combine under one radical.

• \(\dfrac{\sqrt{75}}{\sqrt{3}} = \sqrt{\dfrac{75}{3}} = \sqrt{25}\)
• \(\sqrt{25} = \mathbf{5}\)

Alternatively: \(\sqrt{75} = 5\sqrt{3}\), so \(\dfrac{5\sqrt{3}}{\sqrt{3}} = 5\).

Use \(\sqrt[n]{x} = x^{1/n}\), then apply the power rule \((x^a)^b = x^{ab}\).

• \(\sqrt[5]{x} = x^{1/5}\)
• \(\left(x^{1/5}\right)^3 = x^{(1/5) \cdot 3} = x^{3/5}\)

Denominator (5) = root index. Numerator (3) = outer power.

Use \(x^{m/n} = \left(\sqrt[n]{x}\right)^m\). Take the root first — numbers stay small!

• Denominator 4 → 4th root: \(\sqrt[4]{16} = 2\)  (since \(2^4 = 16\))
• Numerator 3 → cube it: \(2^3 = 8\)
• Result: \(\mathbf{8}\)

Pro tip: Root first keeps numbers manageable. Raising 16 to the 3rd power first gives 4096 — far harder to work with!

• Multiply in numerator (add exponents): \(x^{3/2} \cdot x^{1/2} = x^{4/2} = x^2\)
• Divide (subtract exponents): \(\dfrac{x^2}{x^2} = x^{2-2} = x^0\)
• \(x^0 = \mathbf{1}\)  (any non-zero base to the zero power = 1)

Trap: Don't leave \(x^0\) as your final answer — always simplify it to 1!

The denominator is 5 (odd), so a negative base is allowed in real numbers.

• Take the 5th root first: \(\sqrt[5]{-32} = -2\)  (since \((-2)^5 = -32\))
• Raise to the 3rd power: \((-2)^3 = -8\)
• Result: \(\mathbf{-8}\)

Key check: Is the root index odd or even? Odd → negative base is fine in ℝ.

Apply the power-of-a-product rule \((ab)^n = a^n b^n\), then simplify.

• Expand numerator: \((x^{1/2})^6 \cdot (y^{2/3})^6 = x^{3} \cdot y^{4}\)
• Denominator: \(x^1 \cdot y^3\)
• Divide: \(x^{3-1} \cdot y^{4-3} = x^2 \cdot y = \mathbf{x^2 y}\)