Mathematics • Year 9 • Unit 1 • Lesson 19

Roots, Surds and Real Things

Use fractional indices in real contexts: square paving slabs, the side of a cube of jelly, half-life shrinkage, surd-to-index conversions on a phone-screen diagonal, and a music-string scaling. Then explain in your own words why $x^{1/2}$ has to be $\sqrt{x}$.

Apply · Real-World Maths

1. Word problems

Each problem uses one of the ideas from Lesson 19: $x^{1/n} = \sqrt[n]{x}$, $x^{m/n} = \sqrt[n]{x^m}$, or the index laws applied to fractional indices. Show your working.

1.1 — Paving slab side. A square paving slab has area $144$ cm$^2$.

(a) Use $\text{side} = (\text{area})^{1/2}$ to find the side length.
(b) Briefly explain why the index $1/2$ is the right one to use here. 3 marks

Stuck? "Area is side $\times$ side," so $\text{side}^2 = \text{area}$. The opposite of squaring is the square root, which is the $1/2$ power.

1.2 — Cube of jelly. A cube of jelly has volume $27$ cm$^3$.

(a) Use $\text{side} = (\text{volume})^{1/3}$ to find the side length.
(b) Now find the surface area of one face (side $\times$ side). 3 marks

Stuck? Cube root of $27$ is $3$, because $3 \times 3 \times 3 = 27$. Then square the side for area.

1.3 — TV diagonal in surd-to-index form. A TV screen has width $w = 16$ and height $h = 9$ (in some units). The diagonal length is $d = \sqrt{w^2 + h^2}$.

(a) Calculate $w^2 + h^2$ as a number.
(b) Write the diagonal $d$ in index form (i.e. with a fractional index), then evaluate it to one decimal place. 3 marks

Stuck? Square the two numbers, add, then take the square root — i.e. raise to the $1/2$ power.

1.4 — Half-life shrinkage. An old gadget loses energy by a factor of $\sqrt[3]{8}$ every full week of disuse — i.e. each week the energy is multiplied by $\dfrac{1}{8^{1/3}}$.

(a) Evaluate $8^{1/3}$.
(b) Write the per-week factor $\dfrac{1}{8^{1/3}}$ as a simple fraction.
(c) If the gadget starts at $64$ units of energy, what's left after one week? 3 marks

Stuck on (b)? $\dfrac{1}{8^{1/3}} = \dfrac{1}{\sqrt[3]{8}}$ — find $\sqrt[3]{8}$ first.

1.5 — Big-power evaluation. Evaluate $(\sqrt[3]{x^2})^6$ when $x = 2$. Do this two ways: (a) by simplifying the algebraic expression to a single index first, then substituting; (b) by substituting $x = 2$ first, then evaluating numerically. Confirm both ways give the same answer.

3 marks

Stuck on (a)? $(\sqrt[3]{x^2})^6 = (x^{2/3})^6 = x^{12/3} = x^4$. Way easier than the brute-force second way.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate writes "$16^{1/2} = 8$" because they read the index as "divide by $2$". In your own words, explain (i) what mistake they have made, (ii) what the index $1/2$ actually means, (iii) how the product rule $x^a \cdot x^a = x^{2a}$ forces $x^{1/2}$ to be $\sqrt{x}$, and (iv) the correct value of $16^{1/2}$.

Stuck on (iii)? Try $x^{1/2} \cdot x^{1/2} = x^{1/2 + 1/2} = x^1 = x$. So $x^{1/2}$ is "the number that, multiplied by itself, gives $x$" — that's the definition of $\sqrt{x}$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Paving slab side

(a) Side $= 144^{1/2} = \sqrt{144} = \mathbf{12}$ cm.
(b) The $1/2$ power undoes squaring. The area of a square is $\text{side}^2$, so to recover the side from the area you need the opposite of squaring, which is the square root, i.e. the $1/2$ power.

1.2 — Cube of jelly

(a) Side $= 27^{1/3} = \sqrt[3]{27} = \mathbf{3}$ cm (since $3^3 = 27$).
(b) Area of one face $= 3 \times 3 = \mathbf{9}$ cm$^2$ (or equivalently $3^2 = 27^{2/3} = 9$ using the lesson's formula).

1.3 — TV diagonal

(a) $w^2 + h^2 = 16^2 + 9^2 = 256 + 81 = \mathbf{337}$.
(b) $d = (337)^{1/2} = \sqrt{337} \approx \mathbf{18.4}$ (to 1 d.p.).
Index form uses the $1/2$ power to mean square root — same number, different notation.

1.4 — Half-life shrinkage

(a) $8^{1/3} = \sqrt[3]{8} = \mathbf{2}$.
(b) Per-week factor $= \dfrac{1}{8^{1/3}} = \dfrac{1}{2} = \mathbf{\dfrac{1}{2}}$.
(c) After one week: $64 \times \dfrac{1}{2} = \mathbf{32}$ units of energy.

1.5 — Big-power evaluation

(a) Simplify first: $(\sqrt[3]{x^2})^6 = (x^{2/3})^6 = x^{2/3 \times 6} = x^{12/3} = x^4$. Substitute $x = 2$: $2^4 = \mathbf{16}$.
(b) Substitute first: $\sqrt[3]{2^2} = \sqrt[3]{4} \approx 1.587$, then $(1.587)^6 \approx 16$. ✓
Both ways give $16$ — but way (a) is faster and exact. Simplify before substituting whenever you can.

2.1 — Explain your thinking (sample response)

My classmate has treated the index $1/2$ as if it meant "divide by $2$" — but a fractional index is a root, not a division. The index $1/2$ is shorthand for the square root ($x^{1/n} = \sqrt[n]{x}$, with the denominator giving the root index). The product rule for indices says $x^a \cdot x^a = x^{2a}$, so $x^{1/2} \cdot x^{1/2} = x^{1/2 + 1/2} = x^1 = x$. This means $x^{1/2}$ is the number that, when multiplied by itself, gives $x$ — and that is exactly the definition of $\sqrt{x}$. So $16^{1/2} = \sqrt{16} = \mathbf{4}$, not $8$. A quick check: $4 \times 4 = 16$, so $4$ is the square root; meanwhile $8 \times 8 = 64 \ne 16$, which confirms $8$ cannot be the right answer.

Marking: 1 mark for naming the mistake (treating it as division); 1 for the correct meaning of $1/2$ (square root); 1 for the product-rule argument $x^{1/2} \cdot x^{1/2} = x$; 1 for the correct answer $16^{1/2} = 4$.