Mathematics • Year 9 • Unit 2 • Lesson 19

Solving Quadratics — Mixed Challenge

Six mixed problems mixing trinomial factoring, DOTS, common factor, and the link between roots and $x$-intercepts. One classic DOTS-vs-perfect-square confusion to fix. One open-ended challenge: from roots back to the equation.

Master · Mixed Challenge

1. Mixed problems

Choose the right method (trinomial, DOTS, common factor). Show your working. 3 marks each

1.1 Solve $x^2 + 4x - 21 = 0$ by factoring. Show the factor pair (product, sum) and verify ONE solution by substitution.

1.2 (a) Factor $x^2 - 49$ using DOTS. (b) Hence solve $x^2 - 49 = 0$. (c) State the $x$-intercepts of $y = x^2 - 49$.

1.3 Solve $x^2 = 5x$ by rearranging to $= 0$ and taking out a common factor. State BOTH roots and explain why dividing by $x$ would have lost one.

1.4 Solve $x^2 - 6x + 9 = 0$ by factoring. (Notice anything special about the factor pair?) State why there's only ONE root, and describe what that means for the parabola $y = x^2 - 6x + 9$.

1.5 The parabola $y = x^2 + bx + c$ crosses the $x$-axis at $x = -2$ and $x = 7$. (a) Write the factored form. (b) Expand to find $b$ and $c$. (c) State the axis of symmetry.

1.6 Classify each quadratic by best method to use (DOTS / trinomial factor / common factor), then solve: (a) $x^2 - 36 = 0$, (b) $x^2 - 11x + 30 = 0$, (c) $x^2 + 4x = 0$.

Stuck on 1.6? Look at the structure: $x^2 - $ perfect square? DOTS. Has a constant + middle term? Trinomial. No constant? Common factor.

2. Find the mistake

A student tried to solve $x^2 - 9 = 0$ but confused DOTS with a perfect square. Their working is shown. Identify the error, explain why it's wrong, and write the correct solution. 3 marks

Student's working:

$x^2 - 9 = 0$

Factor: $(x - 3)^2 = 0$ (claims this is DOTS)

So $x - 3 = 0$

Student's final answer: $x = 3$ (one solution).

(a) Identify the mistake.

(b) Verify the mistake by expanding $(x - 3)^2$ and showing it does NOT equal $x^2 - 9$.

(c) Write the correct DOTS factoring of $x^2 - 9$ and state both roots.

Stuck? $(x - 3)^2 = x^2 - 6x + 9$ — three terms, with a middle $x$ term. DOTS gives $(x - 3)(x + 3) = x^2 - 9$ — two terms, no middle.

3. Open-ended challenge — from roots back to the equation

This question has many valid answers. 4 marks

3.1 A parabola has $x$-intercepts at $x = -4$ and $x = 1$, and passes through the point $(0, -8)$.

(a) Write a factored form $y = a(x + 4)(x - 1)$ (note the leading coefficient $a$).
(b) Substitute the point $(0, -8)$ to find $a$.
(c) Expand to write the equation in $y = ax^2 + bx + c$ form.
(d) Verify by solving your final quadratic $= 0$ and checking the roots match $-4$ and $1$.

Stuck? Sub $(0, -8)$: $-8 = a(0 + 4)(0 - 1) = -4a$, so $a = 2$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — $x^2 + 4x - 21 = 0$

Product $-21$, sum $4$: try $7, -3$ ($(7)(-3) = -21$ ✓, $7 + (-3) = 4$ ✓). Factored: $(x + 7)(x - 3) = 0$. Roots: $x = -7$ or $x = 3$. Verify $x = 3$: $9 + 12 - 21 = 0$ ✓.

1.2 — $x^2 - 49 = 0$ (DOTS)

(a) $49 = 7^2$, so $x^2 - 49 = (x - 7)(x + 7)$.
(b) $(x - 7)(x + 7) = 0 \Rightarrow x = 7$ or $x = -7$.
(c) $x$-intercepts of the parabola: $(7, 0)$ and $(-7, 0)$.

1.3 — $x^2 = 5x$

Rearrange: $x^2 - 5x = 0$. Common factor of $x$: $x(x - 5) = 0$. By the null factor law, $x = 0$ or $x = 5$. Two roots. If instead you divide both sides of $x^2 = 5x$ by $x$, you get $x = 5$ — but you've silently assumed $x \neq 0$, throwing away the perfectly valid $x = 0$ root. ALWAYS factor; never divide by a variable.

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

Product $9$, sum $-6$: both equal at $-3$ and $-3$. Factored: $(x - 3)(x - 3) = (x - 3)^2 = 0$. Only ONE root: $x = 3$ (a repeated root). For the parabola $y = x^2 - 6x + 9$, this means the curve TOUCHES the $x$-axis at the vertex $(3, 0)$ but doesn't cross — vertex sits exactly on the $x$-axis.

1.5 — Parabola through $(-2, 0)$ and $(7, 0)$

(a) Factored: $y = (x + 2)(x - 7)$.
(b) Expand: $y = x^2 - 7x + 2x - 14 = x^2 - 5x - 14$. So $b = -5$, $c = -14$.
(c) Axis of symmetry: midpoint of roots, $\dfrac{-2 + 7}{2} = \dfrac{5}{2} = 2.5$. Axis: $x = 2.5$.

1.6 — Pick the method and solve

(a) $x^2 - 36 = 0$: DOTS ($36 = 6^2$). $(x - 6)(x + 6) = 0 \Rightarrow x = \pm 6$.
(b) $x^2 - 11x + 30 = 0$: trinomial factor. Product $30$, sum $-11$: try $-5, -6$. $(x - 5)(x - 6) = 0 \Rightarrow x = 5$ or $x = 6$.
(c) $x^2 + 4x = 0$: common factor ($x$). $x(x + 4) = 0 \Rightarrow x = 0$ or $x = -4$.

2 — Find the mistake (DOTS vs perfect square)

(a) The student wrote $x^2 - 9$ as $(x - 3)^2$, confusing the difference of two squares (DOTS) with a perfect-square trinomial.
(b) Expand $(x - 3)^2 = (x - 3)(x - 3) = x^2 - 3x - 3x + 9 = x^2 - 6x + 9$ — THREE terms, with a middle $-6x$. That does NOT equal $x^2 - 9$ (which is two terms, no middle).
(c) Correct DOTS: $x^2 - 9 = (x - 3)(x + 3) = 0 \Rightarrow x = 3$ or $x = -3$. Two roots: $x = \pm 3$.

3 — Open-ended challenge (sample solution)

(a) Factored: $y = a(x + 4)(x - 1)$.
(b) Sub $(0, -8)$: $-8 = a(0 + 4)(0 - 1) = a(4)(-1) = -4a$. So $a = 2$.
(c) Expand: $y = 2(x + 4)(x - 1) = 2(x^2 + 3x - 4) = 2x^2 + 6x - 8$.
(d) Verify: solve $2x^2 + 6x - 8 = 0 \Rightarrow x^2 + 3x - 4 = 0$ (divide by 2). Product $-4$, sum $3$: try $4, -1$. $(x + 4)(x - 1) = 0 \Rightarrow x = -4$ or $x = 1$. ✓ Roots match.

Marking: 1 mark per correct part (a)/(b)/(c)/(d). Award full marks for the complete construction with verification.