Today's lesson
A quadratic equation can be written in the form $ax^2+bx+c=0$. Today we solve them with algebra — no graphs — by factorising and then using the Null Factor Law.
Learning intentions
- Recognise a quadratic equation and rearrange it into standard form $ax^2+bx+c=0$
- Use the Null Factor Law: if $p\times q=0$, then $p=0$ or $q=0$ (or both)
- Solve by factorising — common factor, monic trinomials, difference of two squares and non-monic trinomials
- Know that a quadratic equation can have 0, 1 or 2 solutions
Part 1 — The Null Factor Law (~8 min)
The whole method rests on one simple fact about multiplying.
Key idea — the Null Factor Law
- If $p\times q=0$, then either $p=0$ or $q=0$ (or both). The only way a product is zero is if a factor is zero.
- So if $(x-3)(x+2)=0$, then $x-3=0$ or $x+2=0$, which gives $x=3$ or $x=-2$.
- To solve a quadratic equation: write it as $ax^2+bx+c=0$, factorise, then set each factor to $0$.
- If every term shares a common factor, divide it out first.
Building understanding — these are already factorised. State the solutions.
- $x(x+1)=0$
- $2x(x-4)=0$
- $(x-3)(x+2)=0$
- $(x+\sqrt3)(x-\sqrt3)=0$
- $(2x-1)(3x+7)=0$
Part 2 — Common factor & difference of two squares (Example 1, ~12 min)
Before factorising a trinomial, always check for a common factor or a difference of two squares ($a^2-b^2=(a-b)(a+b)$).
$\therefore x=0$ or $x-2=0 \;\Rightarrow\; \boxed{x=0 \text{ or } x=2}$
$\therefore \boxed{x=\sqrt{15} \text{ or } x=-\sqrt{15}}$
$\therefore \boxed{x=5 \text{ or } x=-5}$
Now you try: Solve $x^2-3x=0$, $x^2-11=0$, $3x^2=27$. Answers: $x=0$ or $x=3$; $x=\pm\sqrt{11}$; $x=\pm 3$.
📺 Walkthrough: solving $x^2-5x+6=0$ step by step — factorise, then apply the Null Factor Law to read off both solutions.
Part 3 — Monic and non-monic trinomials (Example 2, ~14 min)
For $ax^2+bx+c=0$, factorise the trinomial, then set each factor to zero. When $a=1$ (monic) look for two numbers that multiply to $c$ and add to $b$. When $a\ne 1$ (non-monic) split the middle term.
$\therefore \boxed{x=2 \text{ or } x=3}$
$\therefore \boxed{x=-1}$ (a repeated factor gives only one solution)
$5x(2x-3)+(2x-3)=0 \;\Rightarrow\; (2x-3)(5x+1)=0$
$\therefore 2x=3$ or $5x=-1 \;\Rightarrow\; \boxed{x=\tfrac32 \text{ or } x=-\tfrac15}$
Now you try: Solve $x^2-x-12=0$, $x^2+6x+9=0$, $6x^2+x-2=0$. Answers: $x=-3$ or $x=4$; $x=-3$; $x=-\tfrac23$ or $x=\tfrac12$.
Practice 3.1 — solve by factorising.
- $x^2+7x+10=0$
- $x^2-9x+20=0$
- $x^2+2x-15=0$
- $9x^2-25=0$
- $2x^2+5x-3=0$
- $3x^2-10x+8=0$
b) $x=4$ or $x=5$
c) $x=-5$ or $x=3$
d) $x=\tfrac53$ or $x=-\tfrac53$
e) $x=\tfrac12$ or $x=-3$
f) $x=\tfrac43$ or $x=2$
Part 4 — Disguised quadratics (~8 min)
Some equations don't look like quadratics until you rearrange them into the form $ax^2+bx+c=0$ first.
$\therefore \boxed{x=10 \text{ or } x=-6}$
$\therefore \boxed{x=3 \text{ or } x=-2}$
Now you try: Solve $x^2=2(x+24)$ and $\dfrac{x+20}{x}=x$. Answers: $x=-6$ or $x=8$; $x=-4$ or $x=5$.
Part 5 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. The Null Factor Law says: if $p\times q=0$, then:
Q2. The solutions of $(x-4)(x+2)=0$ are:
Q3. Solving $x^2-7x=0$ gives:
Q4. How many solutions does $x^2+6x+9=0$ have?
Q5. Before solving $2x^2=50$, the best first step is to:
Working program — Cambridge Ex 5G (p452)
After the quiz, open Cambridge Chapter 5 (page 452) and complete the following:
| Set work | Extension |
|---|---|
| Questions 1–7 (½), 8–10 | Questions 11, 12 |
"(½)" means do every second part. Always write the equation as $ax^2+bx+c=0$ before factorising.
Exit ticket — write in your book
- State the Null Factor Law in your own words.
- Why does $x^2+6x+9=0$ have only one solution?
- What is the first thing you check before factorising a trinomial?