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.