Mathematics • Year 9 • Unit 2 • Lesson 19
Solving Quadratics in Real Situations
Four real-world setups that all reduce to factoring a quadratic: a vegetable garden, a flag triangle, a stone dropped from a cliff, a phone-tile dimension puzzle. Set up, factor, solve, then reject roots that don't make physical sense.
1. Word problems
For each: set up the quadratic, factor (or use DOTS), solve, then identify the physically meaningful root. 3 marks each
1.1 — Vegetable garden. A rectangular vegetable garden has length $(x + 4)$ m and width $x$ m. Its area is $32$ m².
(a) Write the area expression as $x(x + 4)$ and set it equal to $32$.
(b) Rearrange to $x^2 + 4x - 32 = 0$ and factor.
(c) Solve and reject the impossible root. State the final dimensions.
1.2 — Flag triangle. A right-angled flag pennant has its short side $x$ cm and its long side $(x + 7)$ cm. Its area is $30$ cm² (recall area of a right-angled triangle $= \tfrac{1}{2} \times $ base $\times$ height).
(a) Write the area equation: $\tfrac{1}{2} \cdot x \cdot (x + 7) = 30$.
(b) Multiply through by $2$ and rearrange to $x^2 + 7x - 60 = 0$. Factor.
(c) Solve and state the short side and long side (cm).
1.3 — Stone from a cliff. A stone is dropped (not thrown) from a cliff of height $80$ m. Its height (m) after $t$ seconds is $h = 80 - 5t^2$.
(a) Set $h = 0$ to model when the stone hits the ground.
(b) Rearrange to $5t^2 = 80$, divide by $5$, then take $\pm \sqrt{\;}$.
(c) Reject the negative root and state the landing time in seconds.
1.4 — Phone tiles. A square phone tile has side $x$ cm. When the side is extended by $5$ cm, the new (larger) square has area $144$ cm².
(a) Write the equation $(x + 5)^2 = 144$.
(b) Take $\pm \sqrt{\;}$ to get $x + 5 = \pm 12$.
(c) Solve and reject the impossible root. State the side length of the original tile.
2. Explain your thinking
Use full sentences. 4 marks
2.1 A classmate solves $(x - 3)(x + 5) = 8$ by setting $x - 3 = 8$ AND $x + 5 = 8$. In your own words, explain (i) why this is wrong, (ii) what condition is required to apply the null factor law correctly, and (iii) write out the correct first step they should have taken.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Vegetable garden
(a) Area equation: $x(x + 4) = 32 \Rightarrow x^2 + 4x = 32$.
(b) Rearrange: $x^2 + 4x - 32 = 0$. Product $-32$, sum $4$: try $8, -4$. Factored: $(x + 8)(x - 4) = 0$.
(c) $x = -8$ or $x = 4$. Reject $x = -8$ (width can't be negative). Accept $x = 4$ m. Dimensions: width $4$ m, length $4 + 4 = 8$ m. Check: $4 \times 8 = 32$ m² ✓.
1.2 — Flag triangle
(a) $\tfrac{1}{2} \cdot x \cdot (x + 7) = 30$.
(b) Multiply by $2$: $x(x + 7) = 60 \Rightarrow x^2 + 7x - 60 = 0$. Product $-60$, sum $7$: try $12, -5$. Factored: $(x + 12)(x - 5) = 0$.
(c) $x = -12$ or $x = 5$. Reject $x = -12$ (length can't be negative). Accept $x = 5$. Short side $= 5$ cm, long side $= 5 + 7 = 12$ cm. Check: $\tfrac{1}{2}(5)(12) = 30$ cm² ✓.
1.3 — Stone from cliff
(a) Set $h = 0$: $80 - 5t^2 = 0$.
(b) $5t^2 = 80 \Rightarrow t^2 = 16 \Rightarrow t = \pm 4$.
(c) Reject $t = -4$ (time can't be negative). Landing time: $t = 4$ seconds.
1.4 — Phone tiles
(a) $(x + 5)^2 = 144$.
(b) Take $\pm \sqrt{\;}$: $x + 5 = \pm 12$, so $x + 5 = 12$ or $x + 5 = -12$.
(c) $x = 7$ or $x = -17$. Reject $x = -17$ (side length must be positive). Original tile side: $x = 7$ cm. (Check: $(7 + 5)^2 = 12^2 = 144$ cm² ✓.)
2.1 — Explain your thinking (sample response)
My classmate's method is wrong because the null factor law only works when the product equals ZERO, not when it equals $8$ (or any other non-zero number). The null factor law says: if $PQ = 0$, then $P = 0$ or $Q = 0$ — this is a special property of zero, not of arbitrary numbers. For $8$, there are many ways to write the product (e.g. $1 \times 8$, $2 \times 4$, $-1 \times -8$, $0.5 \times 16$, etc.), so you cannot conclude what either factor equals just from the product. The correct first step is to EXPAND the brackets and rearrange so that one side is zero: $(x - 3)(x + 5) = 8$ becomes $x^2 + 2x - 15 = 8$, then $x^2 + 2x - 23 = 0$, and now you factor (or use the quadratic formula in later years) on that rearranged form.
Marking: 1 mark for "the law needs zero on one side"; 1 mark for stating the condition ($PQ = 0$); 1 mark for the correct first step (expand + rearrange to $= 0$); 1 mark for clear, full-sentence writing.