Mathematics • Year 9 • Unit 1 • Lesson 19
Fractional Indices
Build fluency with $x^{1/n} = \sqrt[n]{x}$ and $x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$. Read the bottom of the fraction as the root and the top as the power. The "root first, then power" strategy keeps numbers small. All index laws still apply.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Evaluate $27^{2/3}$ without a calculator.
Step 1 — Read the fractional index.
Denominator $3 \to$ cube root. Numerator $2 \to$ square (raise to the power 2).
Reason: $x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$. Bottom of the fraction is the root index; top is the power.
Step 2 — Take the root first.
$\sqrt[3]{27} = 3$
Reason: "root first" keeps the intermediate number tidy. $\sqrt[3]{27^2} = \sqrt[3]{729}$ gives the same answer but a much harder root to take.
Step 3 — Square the result.
$3^2 = 9$
Reason: now apply the numerator power of $2$ to the small number we got from the root.
Step 4 — Check by the other order.
$\sqrt[3]{27^2} = \sqrt[3]{729} = 9$ ✓
Reason: same answer either order — that's the equivalence $x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m$. Root first is just easier arithmetic.
Answer: $\mathbf{27^{2/3} = 9}$.
2. We do — fill in the missing steps
Same structure as Section 1, with the working faded. Fill in each blank. 4 marks
Problem. Evaluate $125^{2/3}$.
Step 1 — Read the fractional index: denominator is __________ , so take the __________ root. Numerator is __________ , so raise to the power __________ .
Step 2 — Take the root first:
$\sqrt[3]{125} = \_\_\_\_$
Step 3 — Apply the power:
$\_\_\_\_^{\,2} = \_\_\_\_\_\_$
Step 4 — Final answer:
$125^{2/3} = \_\_\_\_\_\_$
3. You do — independent practice
Show your working under each problem. The first four are foundation. The middle two are standard. The last two are extension.
Foundation — single root or one rule
3.1 Evaluate $36^{1/2}$. 1 mark
3.2 Evaluate $32^{1/5}$. 1 mark
3.3 Write $\sqrt[3]{x^5}$ in index form. 1 mark
3.4 Simplify $x^{3/4} \cdot x^{1/2}$. 1 mark
Standard — root + power, or two rules
3.5 Evaluate $16^{3/4}$. 2 marks
3.6 Simplify $\dfrac{x^{3/4}}{x^{1/4}}$, leaving the answer in index form. 2 marks
Extension — push your thinking
3.7 Simplify $\left(x^{2/3}\right)^6$, fully. 2 marks
3.8 Simplify $x^{1/2} \cdot x^{1/3}$, giving the answer with a single fractional index. Show your common-denominator working. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $125^{2/3}$)
Step 1: denominator $\mathbf{3}$ $\to$ cube root; numerator $\mathbf{2}$ $\to$ power of $\mathbf{2}$.
Step 2: $\sqrt[3]{125} = \mathbf{5}$.
Step 3: $\mathbf{5}^2 = \mathbf{25}$.
Step 4: $125^{2/3} = \mathbf{25}$.
3.1 — $36^{1/2}$
$36^{1/2} = \sqrt{36} = \mathbf{6}$. (Denominator $2$ is the square root.)
3.2 — $32^{1/5}$
$32^{1/5} = \sqrt[5]{32} = \mathbf{2}$ (since $2^5 = 32$).
3.3 — Write $\sqrt[3]{x^5}$ in index form
Power on top, root on bottom: $\sqrt[3]{x^5} = \mathbf{x^{5/3}}$.
3.4 — $x^{3/4} \cdot x^{1/2}$
Product rule (add indices); common denominator $4$: $\dfrac{3}{4} + \dfrac{1}{2} = \dfrac{3}{4} + \dfrac{2}{4} = \dfrac{5}{4}$. Answer: $\mathbf{x^{5/4}}$.
3.5 — $16^{3/4}$
Root first: $\sqrt[4]{16} = 2$ (since $2^4 = 16$). Then raise to the $3$: $2^3 = \mathbf{8}$.
3.6 — $\dfrac{x^{3/4}}{x^{1/4}}$
Quotient rule (subtract indices): $x^{3/4 - 1/4} = x^{2/4} = \mathbf{x^{1/2}}$ (or equivalently $\sqrt{x}$).
3.7 — $\left(x^{2/3}\right)^6$
Power-of-a-power (multiply indices): $x^{2/3 \times 6} = x^{12/3} = \mathbf{x^4}$. (Simplify $12/3$ to the whole number $4$ — final answer should be in the simplest form possible.)
3.8 — $x^{1/2} \cdot x^{1/3}$
Product rule (add indices). Common denominator $6$: $\dfrac{1}{2} = \dfrac{3}{6}$ and $\dfrac{1}{3} = \dfrac{2}{6}$. Sum: $\dfrac{3}{6} + \dfrac{2}{6} = \dfrac{5}{6}$. Answer: $\mathbf{x^{5/6}}$.
Common slip: writing $x^{2/5}$ by “adding the tops and bottoms” — that's not how fraction addition works.