Definition and Equivalence of Rational Exponents
- Unit Fractional Exponents ($a^{1/n}$):
By the power of a power law:
$$(a^{1/n})^n = a^{(1/n) \times n} = a^1 = a$$
Hence, $a^{1/n}$ is defined as the principal $n^{\text{th}}$ root:
$$a^{1/n} = \sqrt[n]{a}$$
• If $n$ is even, $a$ must be non-negative ($a \ge 0$) for real values.
• If $n$ is odd, roots of negative numbers are real: $(-8)^{1/3} = \sqrt[3]{-8} = -2$. - General Rational Exponent ($a^{m/n}$): $$a^{m/n} = (\sqrt[n]{a})^m = \sqrt[n]{a^m} \quad (n \in \mathbb{Z}^+, m \in \mathbb{Z})$$ Computational Best Practice: Always take the root first to keep numbers small! $$27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9 \quad (\text{preferable to } \sqrt[3]{27^2} = \sqrt[3]{729} = 9)$$
- Negative Rational Exponents: $$a^{-m/n} = \frac{1}{a^{m/n}} = \frac{1}{(\sqrt[n]{a})^m}$$
Example 1 · Criterion A
Evaluating Numerical Expressions with Rational Powers
Evaluate each expression without a calculator:
(a) $16^{3/4}$ (b) $\left(\frac{8}{27}\right)^{-2/3}$ (c) $(-32)^{-3/5}$
- (a) $16^{3/4}$: $$16^{3/4} = (\sqrt[4]{16})^3 = (2)^3 = 8$$
- (b) Inverted negative rational power: $$\left(\frac{8}{27}\right)^{-2/3} = \left(\frac{27}{8}\right)^{2/3} = \left(\sqrt[3]{\frac{27}{8}}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$
- (c) Negative base with odd root: $$(-32)^{-3/5} = \frac{1}{(-32)^{3/5}} = \frac{1}{(\sqrt[5]{-32})^3} = \frac{1}{(-2)^3} = \frac{1}{-8} = -\frac{1}{8}$$
Answers: (a) $8$; (b) $\frac{9}{4}$; (c) $-\frac{1}{8}$.
Simplifying Algebraic Expressions with Fractional Indices
- Applying Index Laws with Fractions: All core laws of indices hold universally for rational exponents $p, q \in \mathbb{Q}$: $$x^p \cdot x^q = x^{p+q}, \quad \frac{x^p}{x^q} = x^{p-q}, \quad (x^p)^q = x^{p \cdot q}$$
- Common Radicand Conversions: $$\sqrt{x} \cdot \sqrt[3]{x} = x^{1/2} \cdot x^{1/3} = x^{1/2 + 1/3} = x^{5/6} = \sqrt[6]{x^5}$$
- Factoring Expressions with Rational Powers: Factor out the power with the smallest exponent: $$x^{3/2} - 4x^{1/2} = x^{1/2}(x^1 - 4) = x^{1/2}(x - 4)$$
Example 2 · Criterion A
Simplifying Mixed Radical and Fractional Expressions
Simplify the algebraic expression completely: $$\frac{\sqrt[3]{8x^4 y^2} \times (x^{1/2} y^{-1/3})^6}{\sqrt{4x^3 y^4}}$$
- Step 1: Convert all radicals into fractional powers: $$\sqrt[3]{8x^4 y^2} = 8^{1/3} x^{4/3} y^{2/3} = 2 x^{4/3} y^{2/3}$$ $$\sqrt{4x^3 y^4} = 4^{1/2} x^{3/2} y^{4/2} = 2 x^{3/2} y^2$$
- Step 2: Expand bracket in numerator: $$(x^{1/2} y^{-1/3})^6 = x^{(1/2) \times 6} y^{(-1/3) \times 6} = x^3 y^{-2}$$
- Step 3: Combine numerator: $$\text{Numerator} = 2 x^{4/3 + 3} y^{2/3 + (-2)} = 2 x^{13/3} y^{-4/3}$$
- Step 4: Divide by denominator: $$\frac{2 x^{13/3} y^{-4/3}}{2 x^{3/2} y^2} = x^{13/3 - 3/2} y^{-4/3 - 2} = x^{26/6 - 9/6} y^{-4/3 - 6/3} = x^{17/6} y^{-10/3} = \frac{x^{17/6}}{y^{10/3}}$$
Answer: $\frac{x^{17/6}}{y^{10/3}}$ or $x^{17/6} y^{-10/3}$.
Solving Radical Equations & Extraneous Solutions
- Radical Equation Method:
1. Isolate the radical term on one side of the equation.
2. Raise both sides to the power matching the root index ($n$).
3. Solve the resulting polynomial (linear or quadratic).
4. MANDATORY: Check for Extraneous Solutions! Squaring both sides can introduce false solutions because $(-k)^2 = (+k)^2$. Substitute all candidate solutions back into the original un-squared equation.
Example 3 · Criterion A & C
Solving a Radical Equation with Extraneous Root Check
Solve the equation $\sqrt{2x + 7} = x - 4$ for $x \in \mathbb{R}$.
- Step 1: Square both sides: $$(\sqrt{2x + 7})^2 = (x - 4)^2$$ $$2x + 7 = x^2 - 8x + 16$$
- Step 2: Form standard quadratic: $$x^2 - 10x + 9 = 0$$ $$(x - 9)(x - 1) = 0 \implies x = 9 \text{ or } x = 1$$
- Step 3: Verification check in original equation:
• Check $x = 9$: $\text{LHS} = \sqrt{2(9) + 7} = \sqrt{25} = 5$. $\text{RHS} = 9 - 4 = 5$. $\text{LHS} = \text{RHS}$ ✓ Valid!
• Check $x = 1$: $\text{LHS} = \sqrt{2(1) + 7} = \sqrt{9} = 3$. $\text{RHS} = 1 - 4 = -3$. $\text{LHS} \ne \text{RHS}$ ✗ Extraneous root!
Answer: $x = 9$ (with $x = 1$ rejected as extraneous).
Convert each radical expression into an equivalent expression in fractional exponent form:
(a) $\sqrt[5]{x^3}$ (b) $\frac{1}{\sqrt[4]{y^7}}$ (c) $\sqrt[3]{\sqrt{z}}$
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(a) $\sqrt[5]{x^3} = x^{3/5}$ [A1]
(b) $\frac{1}{\sqrt[4]{y^7}} = \frac{1}{y^{7/4}} = y^{-7/4}$ [A1]
(c) $\sqrt[3]{\sqrt{z}} = (z^{1/2})^{1/3} = z^{(1/2) \times (1/3)} = z^{1/6}$ [A1]
Evaluate each numerical expression without a calculator:
(a) $81^{3/4}$ (b) $64^{-2/3}$ (c) $\left(\frac{16}{625}\right)^{-3/4}$
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(a) $81^{3/4} = (\sqrt[4]{81})^3 = 3^3 = 27$ [A1]
(b) $64^{-2/3} = \frac{1}{(\sqrt[3]{64})^2} = \frac{1}{4^2} = \frac{1}{16}$ [A1]
(c) $\left(\frac{625}{16}\right)^{3/4} = \left(\frac{\sqrt[4]{625}}{\sqrt[4]{16}}\right)^3 = \left(\frac{5}{2}\right)^3 = \frac{125}{8}$ [A1]
Simplify the algebraic expression: $$\frac{(27 a^6 b^{-3})^{2/3}}{a^{-1} b^4}$$
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Step 1: Distribute exponent $\frac{2}{3}$ to numerator [2 marks]
$27^{2/3} = (\sqrt[3]{27})^2 = 3^2 = 9$ [A1]
$a^{6 \times (2/3)} = a^4$, $b^{-3 \times (2/3)} = b^{-2}$ [A1]
$\text{Numerator} = 9 a^4 b^{-2}$
Step 2: Divide by denominator [2 marks]
$\frac{9 a^4 b^{-2}}{a^{-1} b^4} = 9 a^{4 - (-1)} b^{-2 - 4} = 9 a^5 b^{-6} = \frac{9a^5}{b^6}$ [A2]
Solve the equation for $x \in \mathbb{R}$: $$x^{2/3} = 16$$
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Step 1: Raise both sides to power $\frac{3}{2}$ [2 marks]
$(x^{2/3})^{3/2} = (\pm 16)^{3/2}$
Since the numerator exponent is an even number ($2$), $x$ can be positive or negative [M1].
$x = \pm (\sqrt{16})^3 = \pm 4^3 = \pm 64$ [A2]
Step 2: Verification [1 mark]
$(\pm 64)^{2/3} = ((\pm 64)^{1/3})^2 = (\pm 4)^2 = 16$ ✓ Valid.
$x = \pm 64$ [A1]
Solve the radical equation and state clearly which solutions are valid: $$\sqrt{3x + 1} = x - 3$$
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Step 1: Square both sides [1 mark]
$3x + 1 = (x - 3)^2 = x^2 - 6x + 9$ [M1]
Step 2: Solve quadratic [1 mark]
$x^2 - 9x + 8 = 0 \implies (x - 8)(x - 1) = 0 \implies x = 8 \text{ or } x = 1$ [A1]
Step 3: Check for extraneous roots [2 marks]
• Check $x = 8$: $\sqrt{3(8) + 1} = \sqrt{25} = 5$; $\text{RHS} = 8 - 3 = 5$. True! [A1]
• Check $x = 1$: $\sqrt{3(1) + 1} = \sqrt{4} = 2$; $\text{RHS} = 1 - 3 = -2$. False ($2 \ne -2$). Extraneous! [R1]
Final solution: $\mathbf{x = 8}$.
Solve the equation with fractional powers: $$x^{1/2} - 5 x^{1/4} + 6 = 0$$
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Step 1: Substitution [1 mark]
Let $u = x^{1/4}$. Then $x^{1/2} = (x^{1/4})^2 = u^2$ [M1]:
$u^2 - 5u + 6 = 0$
Step 2: Factorise and solve for $u$ [1 mark]
$(u - 2)(u - 3) = 0 \implies u = 2 \text{ or } u = 3$ [A1]
Step 3: Solve for $x$ [2 marks]
• $x^{1/4} = 2 \implies x = 2^4 = 16$ [A1]
• $x^{1/4} = 3 \implies x = 3^4 = 81$ [A1]
Solutions: $x = 16, 81$.
Solve the equation involving two square root radicals: $$\sqrt{x + 5} - \sqrt{x - 3} = 2$$
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Step 1: Isolate one radical [1 mark]
$\sqrt{x + 5} = 2 + \sqrt{x - 3}$ [M1]
Step 2: Square both sides [1 mark]
$x + 5 = 4 + 4\sqrt{x - 3} + (x - 3) = x + 1 + 4\sqrt{x - 3}$
$x + 5 - (x + 1) = 4\sqrt{x - 3} \implies 4 = 4\sqrt{x - 3} \implies \sqrt{x - 3} = 1$ [M1]
Step 3: Solve and check [2 marks]
$(\sqrt{x - 3})^2 = 1^2 \implies x - 3 = 1 \implies x = 4$ [A1]
Check in original: $\sqrt{4 + 5} - \sqrt{4 - 3} = \sqrt{9} - \sqrt{1} = 3 - 1 = 2$. True! [A1]
Solution: $\mathbf{x = 4}$.
Consider the nested radical expression $y = \sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}}$.
(a) Express the infinite relationship as a quadratic equation in terms of $y$. [2 marks]
(b) Solve the equation for $y$. [2 marks]
(c) Explain why only one of the roots is mathematically possible. [1 mark]
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(a) Equation setup: [2 marks]
Square both sides: $y^2 = 6 + \sqrt{6 + \sqrt{6 + \dots}}$ [M1]
Substitute $y$ for the infinite inner radical: $y^2 = 6 + y \implies y^2 - y - 6 = 0$ [A1]
(b) Solving quadratic: [2 marks]
$(y - 3)(y + 2) = 0 \implies y = 3 \text{ or } y = -2$ [A2]
(c) Physical validity: [1 mark]
Since the principal square root is defined to be non-negative ($y = \sqrt{\dots} \ge 0$), the root $y = -2$ is rejected. Hence, $\mathbf{y = 3}$. [R1]
Johannes Kepler discovered that the orbital period $T$ (in Earth years) of a planet is related to its semi-major axis $r$ (in Astronomical Units, $\text{AU}$) by the power model:
$$T = r^{3/2}$$
(a) Calculate the orbital period of asteroid Ceres, which has an orbital radius of $r = 2.77\text{ AU}$. Round your answer to 3 significant figures. [2 marks]
(b) Dwarf planet Pluto has an orbital period of $T = 248\text{ Earth years}$. Express $r$ in terms of $T$ with a fractional exponent. [1 mark]
(c) Calculate Pluto’s mean orbital radius $r$ in $\text{AU}$ to 3 significant figures. [2 marks]
Show Full Markscheme
(a) Period of Ceres: [2 marks]
$T = (2.77)^{3/2} = (\sqrt{2.77})^3 \approx 4.609\dots \approx \mathbf{4.61\text{ Earth years}}$ [A2]
(b) Inverting formula: [1 mark]
$T = r^{3/2} \implies r = T^{2/3}$ [A1]
(c) Pluto orbital radius: [2 marks]
$r = (248)^{2/3} = (\sqrt[3]{248})^2 \approx (6.28277)^2 \approx \mathbf{39.5\text{ AU}}$ (to 3 s.f.) [A2]
In biological scaling (Kleiber's Law), the basal metabolic rate $B$ (watts) of a mammal scales with body mass $M$ (kg) according to the fractional power model:
$$B = 3.5 \cdot M^{3/4}$$
(a) A human athlete has a body mass of $M = 81\text{ kg}$. Calculate the athlete's basal metabolic rate without a calculator. [2 marks]
(b) If an elephant's mass is $M = 2401\text{ kg}$, calculate its basal metabolic rate without a calculator. [2 marks]
(c) Show that when body mass increases by a factor of $16$, metabolic rate increases by a factor of $8$. [1 mark]
Show Full Markscheme
(a) Human metabolic rate: [2 marks]
$81^{3/4} = (\sqrt[4]{81})^3 = 3^3 = 27$
$B = 3.5 \times 27 = \mathbf{94.5\text{ W}}$ [A2]
(b) Elephant metabolic rate: [2 marks]
$2401 = 7^4 \implies 2401^{3/4} = (\sqrt[4]{2401})^3 = 7^3 = 343$
$B = 3.5 \times 343 = \mathbf{1200.5\text{ W}}$ [A2]
(c) Scaling factor proof: [1 mark]
$$\frac{B(16M)}{B(M)} = \frac{3.5(16M)^{3/4}}{3.5 M^{3/4}} = 16^{3/4} = (\sqrt[4]{16})^3 = 2^3 = 8 \quad \blacksquare$$ [R1]
Fractional Exponents Quick Reference
Rational Index Definitions
Combining Radicals
Power Inversions
Radical Equations Checklist
Top 4 Extended Pitfalls
1. Forgetting Extraneous Root Checks
Squaring $\sqrt{f(x)} = g(x)$ produces solutions where $g(x) < 0$. Since principal roots cannot be negative, you must test all solutions in the original equation to avoid losing Criterion A marks.
2. Missing the $\pm$ on Even Numerators
When solving $x^{2/3} = 25$, students write $x = 25^{3/2} = 125$ and forget that $(-125)^{2/3} = ((-125)^{1/3})^2 = (-5)^2 = 25$. The full solution is $x = \pm 125$.
3. Inverting Negative Exponents Improperly
$a^{-1/2} \ne -a^{1/2}$! Negative exponents represent reciprocals, not negative values: $a^{-1/2} = \frac{1}{\sqrt{a}}$.
4. Squaring Terms Individually
In $\sqrt{x} + 2 = 5$, squaring both sides gives $(\sqrt{x} + 2)^2 = x + 4\sqrt{x} + 4 = 25$, NOT $x + 4 = 25$. Always isolate the radical first: $\sqrt{x} = 3 \implies x = 9$.
Fractional Exponents Check
Answer these 5 questions to test fractional indices, radical conversions, and equations.
What is the exact numerical value of $32^{-2/5}$?
Which single radical expression is equivalent to $\sqrt[3]{x} \times \sqrt{x}$?
Solve for $x$: $x^{-3/4} = 8$.
How many real solutions does the equation $\sqrt{x + 2} = -3$ possess?
If a satellite orbital radius $r$ scales with orbital period $T$ as $r = T^{2/3}$, by what factor does $r$ multiply when $T$ increases by a factor of $27$?