Mathematics • Year 9 • Unit 2 • Lesson 18

Quadratic Equations — Mixed Challenge

Six mixed problems spanning inspection, factoring, the null factor law and the parabola link. One classic null-factor-law misuse to spot. One open-ended challenge: build a quadratic with prescribed roots.

Master · Mixed Challenge

1. Mixed problems

Choose the right method (inspection or factoring), show your working. 3 marks each

1.1 Solve each by inspection: (a) $x^2 = 100$, (b) $x^2 - 144 = 0$, (c) $x^2 + 25 = 0$ (explain why no real solutions).

1.2 Solve $x^2 - 9x + 20 = 0$ by factoring. Show the factor pair (product, sum) and both roots.

1.3 Solve $x^2 + 6x + 9 = 0$ by factoring. (Hint: it's a perfect-square trinomial; the two numbers are equal.) State why there's only ONE solution.

1.4 A parabola has equation $y = x^2 - 4x - 5$. (a) Solve $x^2 - 4x - 5 = 0$ by factoring. (b) State the $x$-intercepts of the parabola as coordinate pairs. (c) State the $y$-intercept (sub $x = 0$).

1.5 Solve $(x + 2)(x - 5) = 0$ using the null factor law. Then expand $(x + 2)(x - 5)$ and write the equation in standard form $x^2 + bx + c = 0$.

1.6 Solve $x^2 = 6x$ by first rearranging to $= 0$, then taking out a common factor of $x$. Show why dividing both sides by $x$ would have LOST a solution.

Stuck on 1.6? $x^2 - 6x = 0 \Rightarrow x(x - 6) = 0 \Rightarrow x = 0$ or $x = 6$. Dividing by $x$ would drop the $x = 0$ solution.

2. Find the mistake

A student tried to solve $(x - 2)(x + 5) = 18$ using the null factor law. Their working is shown. Identify the error, explain why it's wrong, and write the correct solution. 3 marks

Student's working:

$(x - 2)(x + 5) = 18$

Using null factor law: $x - 2 = 18$ and $x + 5 = 18$

So $x = 20$ OR $x = 13$.

Student's final answer: $x = 20$ or $x = 13$.

(a) Identify the mistake.

(b) Explain in one sentence why the null factor law cannot be used directly here.

(c) Show the correct method (expand, rearrange to $= 0$, factor, solve) and state the correct roots.

Stuck? Expand: $x^2 + 3x - 10 = 18 \Rightarrow x^2 + 3x - 28 = 0$. Factor: product $-28$, sum $3$ — try $7$ and $-4$.

3. Open-ended challenge — build the equation

This question has many valid answers. 4 marks

3.1 Construct THREE quadratic equations of the form $x^2 + bx + c = 0$ with the following properties:

(i) An equation whose roots are $x = 1$ and $x = 6$.
(ii) An equation whose roots are $x = -3$ and $x = 4$.
(iii) An equation whose roots are $x = -2$ and $x = -5$ (both negative).

For each: (a) write the factored form, (b) expand to $x^2 + bx + c = 0$, (c) verify by substituting BOTH roots into your standard form.

Stuck? If roots are $r_1$ and $r_2$, the equation is $(x - r_1)(x - r_2) = 0$. Expand to get $x^2 - (r_1 + r_2)x + r_1 r_2 = 0$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Three inspection problems

(a) $x = \pm 10$.
(b) $x^2 = 144 \Rightarrow x = \pm 12$.
(c) $x^2 = -25$ has NO real solutions because no real number squared can give a negative result; the parabola $y = x^2 + 25$ has vertex $(0, 25)$ and never crosses the $x$-axis.

1.2 — $x^2 - 9x + 20 = 0$

Product $20$, sum $-9$: both negative. Try $-4$ and $-5$ ($(-4)(-5) = 20$ ✓, $-4 + (-5) = -9$ ✓). Factored: $(x - 4)(x - 5) = 0$. Roots: $x = 4$ or $x = 5$.

1.3 — $x^2 + 6x + 9 = 0$ (perfect square)

Product $9$, sum $6$: both equal at $3$ and $3$. Factored: $(x + 3)(x + 3) = (x + 3)^2 = 0$. So $x + 3 = 0 \Rightarrow x = -3$. Only ONE solution because both factors give the SAME root — the parabola $y = x^2 + 6x + 9$ just touches the $x$-axis at the vertex $(-3, 0)$, not crossing it.

1.4 — Parabola $y = x^2 - 4x - 5$

(a) Product $-5$, sum $-4$: try $-5$ and $1$. Factored: $(x - 5)(x + 1) = 0 \Rightarrow x = 5$ or $x = -1$.
(b) $x$-intercepts: $(5, 0)$ and $(-1, 0)$.
(c) $y$-intercept: sub $x = 0$: $y = 0 - 0 - 5 = -5$. Point $(0, -5)$.

1.5 — $(x + 2)(x - 5) = 0$

Null factor law: $x + 2 = 0$ or $x - 5 = 0$, so $x = -2$ or $x = 5$. Expand: $(x + 2)(x - 5) = x^2 - 5x + 2x - 10 = x^2 - 3x - 10$. Standard form: $x^2 - 3x - 10 = 0$.

1.6 — $x^2 = 6x$ vs dividing by $x$

Rearrange: $x^2 - 6x = 0 \Rightarrow x(x - 6) = 0 \Rightarrow x = 0$ or $x = 6$. (Two solutions.) If you instead divide both sides of $x^2 = 6x$ by $x$, you get $x = 6$ — only ONE solution. The problem is that division by $x$ assumes $x \neq 0$, which silently throws away the perfectly valid $x = 0$ root. ALWAYS rearrange to $= 0$ and factor; never divide by a variable.

2 — Find the mistake (null factor misuse)

(a) The student applied the null factor law to a product that equals $18$, NOT $0$.
(b) The null factor law states: if a product $PQ = 0$, then $P = 0$ or $Q = 0$. It only works when the product is ZERO — you can't conclude that one factor equals $18$ just because the product does (lots of pairs multiply to $18$: $2 \times 9$, $3 \times 6$, $18 \times 1$, etc.).
(c) Correct method: expand $(x - 2)(x + 5) = x^2 + 3x - 10$. Set $= 18$: $x^2 + 3x - 10 = 18 \Rightarrow x^2 + 3x - 28 = 0$. Factor (product $-28$, sum $3$): try $7$ and $-4$, $(x + 7)(x - 4) = 0$. Roots: $x = -7$ or $x = 4$. (Quick check: $(4 - 2)(4 + 5) = 2 \times 9 = 18$ ✓.)

3 — Open-ended challenge (sample solutions)

(i) Roots $1, 6$. Factored: $(x - 1)(x - 6) = 0$. Expand: $x^2 - 7x + 6 = 0$. Check $x = 1$: $1 - 7 + 6 = 0$ ✓. Check $x = 6$: $36 - 42 + 6 = 0$ ✓.

(ii) Roots $-3, 4$. Factored: $(x + 3)(x - 4) = 0$. Expand: $x^2 - x - 12 = 0$. Check $x = -3$: $9 + 3 - 12 = 0$ ✓. Check $x = 4$: $16 - 4 - 12 = 0$ ✓.

(iii) Roots $-2, -5$. Factored: $(x + 2)(x + 5) = 0$. Expand: $x^2 + 7x + 10 = 0$. Check $x = -2$: $4 - 14 + 10 = 0$ ✓. Check $x = -5$: $25 - 35 + 10 = 0$ ✓.

Marking: 1 mark per correct factored + expanded form with verified roots, plus 1 mark for clear, organised presentation. Award full marks for any three correct, verified equations.