Mathematics • Year 9 • Unit 2 • Lesson 19
Solving Quadratic Equations
Drill three solving routes for $x^2 + bx + c = 0$: factor a trinomial (product/sum trick), spot a difference of squares (DOTS), or take out a common factor. Always finish with the null factor law and a substitution check.
1. I do — fully worked example (factor a trinomial)
Read every line. Each step shows why we choose that move.
Problem. Solve $x^2 - 5x + 6 = 0$ by factoring. State the $x$-intercepts of $y = x^2 - 5x + 6$.
Step 1 — Read $b$ and $c$.
$b = -5$, $c = +6$. Both signs matter: $c > 0$ means BOTH factor numbers have the same sign; $b < 0$ means both are NEGATIVE.
Reason: sign rules tell you what sort of factor pair to look for before you start guessing.
Step 2 — Find the two numbers.
Product $= 6$, sum $= -5$. Try $-2, -3$: $(-2)(-3) = 6$ ✓; $(-2) + (-3) = -5$ ✓.
Reason: "product = $c$, sum = $b$" is the universal trinomial trick.
Step 3 — Write the factored form.
$x^2 - 5x + 6 = (x - 2)(x - 3)$. So the equation becomes $(x - 2)(x - 3) = 0$.
Reason: signs match the chosen numbers — both NEGATIVE, so both factors $(x - \ldots)$.
Step 4 — Apply the null factor law.
$x - 2 = 0$ OR $x - 3 = 0$, so $x = 2$ or $x = 3$.
Reason: a product of two factors is zero only when at least one factor is zero.
Step 5 — Substitute back to check.
$x = 2$: $4 - 10 + 6 = 0$ ✓. $x = 3$: $9 - 15 + 6 = 0$ ✓. Both roots verified.
Reason: substitution catches algebra errors. $x$-intercepts of the parabola: $(2, 0)$ and $(3, 0)$ — same numbers as the roots.
Answer: $x = 2$ or $x = 3$; $x$-intercepts $(2, 0)$ and $(3, 0)$.
2. We do — fill in the missing steps (DOTS)
Same routine but with the working faded. Fill in each blank. 4 marks
Problem. Solve $x^2 - 16 = 0$ using the difference of two squares (DOTS).
Step 1 — Recognise DOTS: the expression is a DIFFERENCE of two terms; the second term is a perfect square ($16 = \_\_^2$).
Step 2 — Apply $a^2 - b^2 = (a - b)(a + b)$: $x^2 - 16 = (x - \_\_\_)(x + \_\_\_)$.
Step 3 — Use the null factor law: $(x - 4)(x + 4) = 0 \Rightarrow x - 4 = 0$ OR $x + 4 = 0$.
Step 4 — Solve and check: $x = \_\_\_$ or $x = \_\_\_$. Quick check for $x = 4$: $4^2 - 16 = \_\_\_$ ✓.
3. You do — independent practice
Show your working. The first four are foundation (clean trinomials and DOTS). The middle two are standard (sign awareness and common factor). The last two are extension (link to the parabola, no-real-roots case).
Foundation — clean factor drills
3.1 Solve $x^2 + 7x + 12 = 0$ by factoring. 1 mark
3.2 Solve $x^2 - 25 = 0$ using DOTS. 1 mark
3.3 Solve $x^2 - 8x + 15 = 0$ by factoring. 1 mark
3.4 Solve $x^2 - 100 = 0$ using DOTS. 1 mark
Standard — sign awareness and common factor
3.5 Solve $x^2 - x - 6 = 0$ (opposite-sign factor pair). Show product, sum, factored form, both roots. 2 marks
3.6 Solve $x^2 - 8x = 0$ by taking out a common factor of $x$. State both roots and explain why dividing by $x$ would lose one. 2 marks
Extension — link to the parabola, or no-real-solution
3.7 (a) Solve $x^2 + 2x - 15 = 0$ by factoring. (b) State the $x$-intercepts of the parabola $y = x^2 + 2x - 15$ as coordinate pairs. (c) Find the $y$-intercept by substituting $x = 0$. 3 marks
3.8 (a) Try to factor $x^2 + 9$ using DOTS — explain why it doesn't work. (b) Solve $x^2 + 9 = 0$ by inspection. (c) Sketch a brief description of why the parabola $y = x^2 + 9$ has NO $x$-intercepts. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $x^2 - 16 = 0$)
Step 1: $16 = \mathbf{4}^2$.
Step 2: $x^2 - 16 = (x - \mathbf{4})(x + \mathbf{4})$.
Step 3: null factor law gives $x - 4 = 0$ or $x + 4 = 0$.
Step 4: $x = \mathbf{4}$ or $x = \mathbf{-4}$. Check $x = 4$: $16 - 16 = \mathbf{0}$ ✓.
3.1 — $x^2 + 7x + 12 = 0$
Product $12$, sum $7$: try $3, 4$. $(x + 3)(x + 4) = 0 \Rightarrow x = -3$ or $x = -4$.
3.2 — $x^2 - 25 = 0$
$25 = 5^2$. DOTS: $(x - 5)(x + 5) = 0 \Rightarrow x = \pm 5$.
3.3 — $x^2 - 8x + 15 = 0$
Product $15$, sum $-8$: both negative, try $-3, -5$. $(x - 3)(x - 5) = 0 \Rightarrow x = 3$ or $x = 5$.
3.4 — $x^2 - 100 = 0$
$100 = 10^2$. DOTS: $(x - 10)(x + 10) = 0 \Rightarrow x = \pm 10$.
3.5 — $x^2 - x - 6 = 0$
Product $-6$, sum $-1$: opposite signs (because product is negative). Try $-3, 2$ ($(-3)(2) = -6$ ✓, $-3 + 2 = -1$ ✓). Factored: $(x - 3)(x + 2) = 0$. Roots: $x = 3$ or $x = -2$.
3.6 — $x^2 - 8x = 0$
Common factor of $x$: $x(x - 8) = 0$. By the null factor law, $x = 0$ or $x - 8 = 0$, so $x = 0$ or $x = 8$. If you divide both sides by $x$, you get $x - 8 = 0 \Rightarrow x = 8$ — but you've lost the $x = 0$ solution! Dividing by $x$ silently assumes $x \neq 0$, which throws away a valid root. ALWAYS factor; never divide by a variable.
3.7 — $x^2 + 2x - 15 = 0$
(a) Product $-15$, sum $2$: try $5, -3$ ($(5)(-3) = -15$ ✓, $5 + (-3) = 2$ ✓). Factored: $(x + 5)(x - 3) = 0 \Rightarrow x = -5$ or $x = 3$.
(b) $x$-intercepts of the parabola: $(-5, 0)$ and $(3, 0)$.
(c) $y$-intercept: $y = 0 + 0 - 15 = -15$. Point $(0, -15)$.
3.8 — $x^2 + 9 = 0$
(a) DOTS is $a^2 - b^2 = (a - b)(a + b)$. $x^2 + 9$ is a SUM of squares, not a difference, so DOTS doesn't apply — there are no real $a, b$ such that $(x + a)(x + b)$ expands to $x^2 + 9$ with no $x$-term and $ab = 9$ requires same signs, but then $a + b \neq 0$.
(b) Inspection: $x^2 = -9$. No real number squared can give $-9$, so there are NO real solutions.
(c) The parabola $y = x^2 + 9$ has vertex $(0, 9)$ and opens upward — its lowest point is $9$ above the $x$-axis, so it never crosses, confirming no real $x$-intercepts.