Mathematics • Year 9 • Unit 1 • Lesson 2
Evaluating Powers — Mixed Challenge
Pull together everything from Unit 1 so far: index notation (L1), evaluating positive, negative and fractional powers (L2). You'll choose the right approach, spot a sign-mistake, and tackle an open-ended puzzle.
1. Mixed problems — choose the right approach
Each question uses a different combination of ideas from Lessons 1 and 2. Decide which rule applies before you start writing. Show your working. 3 marks each
1.1 Evaluate $(-5)^3$ and $-5^3$. Show that they give the same answer, and explain in one sentence why.
1.2 Evaluate $\left(\dfrac{3}{5}\right)^3$ as a fraction in lowest terms.
1.3 Use BIDMAS to evaluate $\;30 - 4 \times 2^2 + 5$. Show every step.
1.4 Evaluate $(-1)^7 \times (-2)^4$. Predict the sign first, then evaluate.
1.5 Without using a calculator, decide which is bigger: $\;(-2)^6\;$ or $\;-3^4$. Show both values.
1.6 Evaluate $\;\left(\dfrac{1}{2}\right)^3 + \left(\dfrac{1}{2}\right)^2\;$ as a single fraction in lowest terms.
2. Find the mistake
Another student has tried to evaluate $\;(-2)^4$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain why it's wrong, then re-do the working correctly. 3 marks
Student's working — evaluate $(-2)^4$:
Line 1: $(-2)^4 = (-2) \times (-2) \times (-2) \times (-2)$
Line 2: $= -(2 \times 2 \times 2 \times 2)$
Line 3: $= -(16)$
Line 4: So $(-2)^4 = -16$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? Pair up the negatives: $(-2) \times (-2) = +4$, not $-4$. Even number of negatives multiplied gives a positive.3. Open-ended challenge — Same expression, two values
This question has multiple valid answers — there are several different bases that work. 4 marks
3.1 Find two different positive whole-number bases $a$ and $b$ (with $a \ne b$, both at least $2$) such that
$\;\;(-a)^3 + b^2 = 0$.
For each pair you find:
(i) Write down $a$ and $b$.
(ii) Evaluate $(-a)^3$ and $b^2$ to confirm they cancel.
(iii) Explain in one sentence why an odd index on the negative is essential.
Bonus: Try to find at least two distinct pairs $(a, b)$.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $(-5)^3$ vs $-5^3$
$(-5)^3 = (-5)(-5)(-5) = (25)(-5) = \mathbf{-125}$.
$-5^3 = -(5 \times 5 \times 5) = -(125) = \mathbf{-125}$.
Both equal $-125$. Why the same: odd indices preserve the sign — when the index is odd, $(-a)^n$ and $-a^n$ give the same value.
1.2 — $(3/5)^3$
$\dfrac{3^3}{5^3} = \mathbf{\dfrac{27}{125}}$ (already in lowest terms — gcd$(27,125) = 1$).
1.3 — $30 - 4 \times 2^2 + 5$
BIDMAS: $2^2 = 4$; then $4 \times 4 = 16$; then $30 - 16 + 5 = 14 + 5 = \mathbf{19}$.
1.4 — $(-1)^7 \times (-2)^4$
Prediction: $(-1)^7 = -1$ (odd index), $(-2)^4 = +16$ (even index). So the product is negative.
Evaluation: $-1 \times 16 = \mathbf{-16}$.
1.5 — Compare $(-2)^6$ and $-3^4$
$(-2)^6 = +64$ (even index of negative base).
$-3^4 = -(3^4) = -(81) = -81$.
So $\mathbf{(-2)^6 > -3^4}$ (i.e. $64 > -81$ — a positive number is always bigger than a negative).
1.6 — $(1/2)^3 + (1/2)^2$
$\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}$. $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.
Common denominator $8$: $\dfrac{1}{8} + \dfrac{2}{8} = \mathbf{\dfrac{3}{8}}$.
2 — Find the mistake
(a) The mistake is on Line 2.
(b) The student has factored out a single negative sign as if there were an odd number of negatives, but there are four negatives — pairs of negatives multiply to positives, so $(-2) \times (-2) \times (-2) \times (-2) = +(2 \times 2 \times 2 \times 2)$, not $-(2 \times 2 \times 2 \times 2)$. Even index of a negative base gives a positive result.
(c) Corrected working:
$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2)$
$= (+4) \times (+4)$
$= +16$
$= \mathbf{16}$.
This is the trap from the lesson's "Spot the Trap" card — students often forget that pairs of negatives cancel.
3 — Open-ended (sample solution)
We need $(-a)^3 + b^2 = 0$, i.e. $b^2 = a^3$. (Note $(-a)^3 = -a^3$ since the index is odd.)
Pair 1: $a = 4$, $b = 8$.
$(-4)^3 = -64$. $8^2 = 64$. Sum $= -64 + 64 = 0$ ✓.
Why odd index matters: if the index on $-a$ were even, $(-a)^3$ would be positive, and adding $b^2$ (also positive) could never give $0$.
Pair 2: $a = 9$, $b = 27$.
$(-9)^3 = -729$. $27^2 = 729$. Sum $= -729 + 729 = 0$ ✓.
Other valid approaches: Any pair with $b^2 = a^3$, like $a = 16, b = 64$ ($64^2 = 4096 = 16^3$ ✓), or in general $a = k^2$ and $b = k^3$ for any whole $k \ge 2$.
Marking: 2 marks per valid pair (one for the pair, one for the working confirming $0$). Up to 4 total. 1 bonus mark if the odd-index reasoning is clearly stated.