EQUATIONS

QUESTION 1 [4 marks] — Criterion A Medium

State the inverse operation needed to undo each of the following, then use it to solve for $x$:

a. $x+12=30$

[1]

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Inverse of $+12$ is $-12$: $$x=30-12=18$$

b. $x-9=4$

[1]

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Inverse of $-9$ is $+9$: $$x=4+9=13$$

c. $7x=63$

[1]

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Inverse of $\times7$ is $\div7$: $$x=63\div7=9$$

d. $\dfrac{x}{6}=3$

[1]

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Inverse of $\div6$ is $\times6$: $$x=3\times6=18$$

QUESTION 2 [4 marks] — Criterion A Medium

Solve $\sqrt{x}=7$ and $x^2=49$ (for $x>0$), using inverse operations.

a. State the inverse of 'square root' and use it to solve $\sqrt{x}=7$.

[2]

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Inverse of $\sqrt{\ }$ is squaring: $$x=7^2=49$$

b. State the inverse of 'squaring' and use it to solve $x^2=49$ for $x>0$.

[2]

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Inverse of squaring is square-rooting: $$x=\sqrt{49}=7$$

QUESTION 3 [5 marks] — Criterion B Medium

Investigate why squaring and square-rooting are inverse operations, and why order matters when applying two inverse operations in sequence.

a. Start with $x=5$. Square it, then take the square root of the result. What do you get?

[2]

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$5^2=25$, then $\sqrt{25}=5$ — back to the original 5.

b. Now start with $x=5$ again. Take the square root FIRST (of 5, giving an irrational number), then square that result. Do you still get back to 5?

[2]

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$\sqrt{5}\approx2.236$, then $(\sqrt{5})^2=5$ — yes, still returns to 5 (since squaring exactly undoes square-rooting, in either order, as long as you stay with the SAME operations as a pair).

c. Explain why applying an operation and then its EXACT inverse always 'undoes' the original change, regardless of the order.

[1]

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By definition, an inverse operation is specifically designed to exactly reverse another operation — applying an operation then its inverse (or vice versa) always returns you to the starting value, since that's precisely what 'inverse' means.

QUESTION 4 [4 marks] — Criterion C Medium

A student solving $x-4=10$ uses the WRONG inverse operation, writing '$x=10-4=6$' (subtracting again instead of adding).

a. Explain the student's error, and identify the CORRECT inverse operation needed.

[2]

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The equation has $-4$ applied to $x$; to undo (invert) subtraction, you must ADD, not subtract again. Correct: $x=10+4=14$.

b. Verify the correct answer, and show why the student's answer ($x=6$) fails when substituted back into the original equation.

[2]

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Correct check: $14-4=10$ ?. Student's check: $6-4=2\ne10$ ? — confirms $x=6$ is wrong.

QUESTION 5 [4 marks] — Criterion D Medium

A hot air balloon is at height $h$ metres. After descending 35 m, it is at 60 m.

a. Write an equation for this situation ($h-35=60$), and use the correct inverse operation to solve for $h$.

[2]

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Inverse of $-35$ is $+35$: $$h=60+35=95 \text{ m}$$

b. The balloon then needs to rise back to its original height of 95 m from 60 m, at a constant rate of 5 m per minute. Using inverse operations (or otherwise), find how long this will take.

[2]

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Height still needed: $95-60=35$ m. Time $=35\div5=7$ minutes.

QUESTION 6 [4 marks] — Criterion A Medium

Use inverse operations to solve each equation:

a. $x^3=125$

[2]

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Inverse of cubing is cube-rooting: $$x=\sqrt[3]{125}=5$$

b. $\sqrt{x}=8$

[2]

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Inverse of square-rooting is squaring: $$x=8^2=64$$

QUESTION 7 [4 marks] — Criterion A Medium

Solve $2\sqrt{x}+3=15$, using inverse operations in the CORRECT order.

a. Isolate the square root term first (undo the $+3$, then undo the $\times2$).

[3]

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$2\sqrt{x}=12 \Rightarrow \sqrt{x}=6$.

b. Complete the solution by undoing the square root.

[1]

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$$x=6^2=36$$

QUESTION 8 [5 marks] — Criterion B Hard

Investigate whether 'squaring then square-rooting' and 'square-rooting then squaring' ALWAYS return you to the original number, testing with a NEGATIVE starting value, $x=-4$.

a. Square $-4$, then take the square root of the result. Do you return to $-4$?

[2]

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$(-4)^2=16$. $\sqrt{16}=4$ — this does NOT return to $-4$; it gives $+4$ instead.

b. Explain WHY this happens, referring to the fact that the square root symbol $\sqrt{\ }$ conventionally refers to the POSITIVE root only, even when the original number being squared was negative.

[3]

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The square root operation, by mathematical CONVENTION, always returns the non-negative (principal) root — so $\sqrt{16}=4$ specifically, never $-4$, even though BOTH $4^2$ and $(-4)^2$ equal 16. This means 'square then square-root' does NOT perfectly reverse for negative starting numbers — the true inverse relationship only holds cleanly for non-negative starting values, which is an important subtlety when using square-rooting as an 'inverse' of squaring.

QUESTION 9 [3 marks] — Criterion C Medium

A student solving $x^2=49$ using inverse operations writes only '$x=7$', missing a second valid solution.

a. Explain the student's error, and state BOTH correct solutions.

[3]

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Squaring is NOT a one-to-one operation — both $7^2=49$ AND $(-7)^2=49$, so 'undoing' a square must consider BOTH the positive and negative square roots. Correct solutions: $x=7$ or $x=-7$ (i.e. $x=\pm7$).

QUESTION 10 [6 marks] — Criterion D Hard

The relationship between a cube's volume $V$ and its side length $s$ is $V=s^3$. A storage cube has volume $V=343$ cm³.

a. Use the inverse operation (cube root) to find the side length.

[2]

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$$s=\sqrt[3]{343}=7\text{ cm}$$

b. A larger storage cube has EXACTLY DOUBLE the volume of this one (686 cm³). Find its side length (to 2 decimal places), and determine whether DOUBLING the volume also DOUBLES the side length (compare the new side length to double the original 7cm).

[4]

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$s=\sqrt[3]{686}\approx8.82$cm. This is NOT double the original side length (which would be $14$cm) — doubling the VOLUME of a cube does NOT double its side length, since volume scales with the CUBE of the side length, not linearly; a much smaller increase in side length (from 7cm to about 8.82cm) is enough to double the volume.

QUESTION 11 [6 marks] — Criterion D Hard

A physics formula relates a pendulum's period $T$ (seconds) to its length $L$ (metres): $T=2\pi\sqrt{\dfrac{L}{9.8}}$. A pendulum has a measured period of $T=2$ seconds.

a. Rearrange the formula (using inverse operations step by step) to solve for $L$ in terms of $T$.

[4]

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$\dfrac{T}{2\pi}=\sqrt{\dfrac{L}{9.8}}$ (divide by $2\pi$). $\left(\dfrac{T}{2\pi}\right)^2=\dfrac{L}{9.8}$ (square both sides). $$L=9.8\left(\frac{T}{2\pi}\right)^2$$

b. Substitute $T=2$ to find the pendulum's length, correct to 2 decimal places.

[2]

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$$L=9.8\left(\frac{2}{2\pi}\right)^2=9.8\times\left(\frac{1}{\pi}\right)^2\approx0.99\text{ m}$$

Inverse operations