Mathematics • Year 9 • Unit 2 • Lesson 18
Introduction to Quadratic Equations
Build the two routes for solving $ax^2 + bx + c = 0$: inspection (no $bx$ term, take $\pm \sqrt{\;}$) and factoring (use the null factor law). One worked, one guided, eight independent.
1. I do — fully worked example (inspection)
Read every line. Each step shows why we take the step, not just what the result is.
Problem. Solve $x^2 - 9 = 0$ by inspection. Check using the parabola $y = x^2 - 9$.
Step 1 — Recognise the form.
No $bx$ term ($b = 0$). Just $x^2$ and a constant. Inspection is the fastest route.
Reason: when there's no middle $x$ term, you don't need factoring — just isolate $x^2$.
Step 2 — Isolate $x^2$.
$x^2 - 9 = 0 \Rightarrow x^2 = 9$.
Reason: add $9$ to both sides.
Step 3 — Take $\pm \sqrt{\;}$ of both sides.
$x = \pm \sqrt{9} = \pm 3$. Two solutions: $x = 3$ AND $x = -3$.
Reason: both $3^2 = 9$ AND $(-3)^2 = 9$. Always include the $\pm$.
Step 4 — Check using the parabola.
$y = x^2 - 9$ has vertex $(0, -9)$ and crosses the $x$-axis at $x = \pm 3$.
Reason: the $x$-intercepts of the parabola ARE the roots of the equation. Two views, same answer.
Answer: $x = \pm 3$.
2. We do — fill in the missing steps (factoring)
Same routine but with the working faded. Fill in each blank. 4 marks
Problem. Solve $x^2 - 5x + 6 = 0$ by factoring.
Step 1 — Find two numbers that MULTIPLY to $c = \_\_\_$ AND ADD to $b = \_\_\_$. Try $-2$ and $-3$: $(-2)(-3) = \_\_\_$ ✓ and $(-2) + (-3) = \_\_\_$ ✓.
Step 2 — Write the factored form: $x^2 - 5x + 6 = (x \_\_\_)(x \_\_\_) = 0$.
Step 3 — Apply the null factor law: if $PQ = 0$ then $P = 0$ or $Q = 0$. So $x - 2 = 0$ OR $x - 3 = 0$.
Step 4 — Solve each: $x = \_\_\_$ or $x = \_\_\_$. Check $x = 2$: $4 - 10 + 6 = \_\_\_$ ✓.
3. You do — independent practice
Show your working. The first four are foundation (one method). The middle two are standard (choose method, both roots). The last two are extension (link to the parabola or no-solution case).
Foundation — single-method drills
3.1 Solve $x^2 = 25$ by inspection. 1 mark
3.2 Solve $x^2 - 16 = 0$ by inspection. 1 mark
3.3 Solve $(x - 2)(x - 3) = 0$ using the null factor law. 1 mark
3.4 Solve $(x + 5)(x - 4) = 0$ using the null factor law. 1 mark
Standard — find the factor pair, solve, state both roots
3.5 Solve $x^2 - 7x + 12 = 0$ by factoring. Show the two numbers (product $12$, sum $-7$), the factored form, and both roots. 2 marks
3.6 Solve $x^2 + 3x - 10 = 0$ by factoring. (Note: product is negative — that means opposite signs in the factor pair.) 2 marks
Extension — link to the parabola, or the no-solution case
3.7 Solve $x^2 + 2x - 8 = 0$ by factoring, then state the $x$-intercepts of the parabola $y = x^2 + 2x - 8$ (as coordinate pairs). 3 marks
3.8 Solve each by inspection: (a) $x^2 = 49$, (b) $x^2 = 0$, (c) $x^2 + 4 = 0$. For (c), explain in one sentence why there are no real solutions. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $x^2 - 5x + 6 = 0$)
Step 1: $c = \mathbf{6}$, $b = \mathbf{-5}$. $(-2)(-3) = \mathbf{6}$ ✓; $(-2) + (-3) = \mathbf{-5}$ ✓.
Step 2: $(x \mathbf{- 2})(x \mathbf{- 3}) = 0$.
Step 3: null factor law gives $x - 2 = 0$ OR $x - 3 = 0$.
Step 4: $x = \mathbf{2}$ or $x = \mathbf{3}$. Check $x = 2$: $4 - 10 + 6 = \mathbf{0}$ ✓.
3.1 — $x^2 = 25$
$x = \pm \sqrt{25} = \pm 5$. Both $5$ and $-5$ work.
3.2 — $x^2 - 16 = 0$
$x^2 = 16 \Rightarrow x = \pm 4$.
3.3 — $(x - 2)(x - 3) = 0$
$x - 2 = 0$ or $x - 3 = 0$, so $x = 2$ or $x = 3$.
3.4 — $(x + 5)(x - 4) = 0$
$x + 5 = 0$ or $x - 4 = 0$, so $x = -5$ or $x = 4$. (Sign FLIPS: $(x + 5) = 0$ means $x = -5$.)
3.5 — $x^2 - 7x + 12 = 0$
Product $12$, sum $-7$: choose $-3$ and $-4$ ($(-3)(-4) = 12$ ✓, $-3 + (-4) = -7$ ✓). Factored: $(x - 3)(x - 4) = 0$. Roots: $x = 3$ or $x = 4$.
3.6 — $x^2 + 3x - 10 = 0$
Product $-10$, sum $3$: opposite signs. Try $5$ and $-2$ ($(5)(-2) = -10$ ✓, $5 + (-2) = 3$ ✓). Factored: $(x + 5)(x - 2) = 0$. Roots: $x = -5$ or $x = 2$.
3.7 — $x^2 + 2x - 8 = 0$
Product $-8$, sum $2$: try $4$ and $-2$ ($(4)(-2) = -8$ ✓, $4 + (-2) = 2$ ✓). Factored: $(x + 4)(x - 2) = 0$. Roots: $x = -4$ or $x = 2$. $x$-intercepts of the parabola: $(-4, 0)$ and $(2, 0)$ — same numbers, written as coordinate pairs.
3.8 — Three inspection problems
(a) $x = \pm 7$.
(b) $x^2 = 0 \Rightarrow x = 0$ (one solution — the parabola touches the $x$-axis at the vertex).
(c) $x^2 = -4$ has NO real solutions because no real number squared can give a negative result: $0^2 = 0$, positive$^2 > 0$, and negative$^2 > 0$ too. The parabola $y = x^2 + 4$ sits entirely above the $x$-axis (vertex $(0, 4)$), so it never crosses.