Lesson 19 ~25 min Unit 2 · Non-Linear +85 XP

Solving Quadratic Equations

Set $y = 0$. Factor. Use the null factor law. The two solutions are the $x$-intercepts of the parabola — algebra and geometry, one story.

Today's hook: $x^2 - 5x + 6 = 0$. Can you find the two values of $x$ without graphing? And how do those values appear on the parabola $y = x^2 - 5x + 6$?

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Think First

warm-up

Imagine the parabola $y = x^2 - 5x + 6$. You've sketched dozens already. Today you'll find its $x$-intercepts WITHOUT plotting — just by solving $x^2 - 5x + 6 = 0$. Try it: what two numbers multiply to $6$ and add to $-5$? Write them, then write $x^2 - 5x + 6$ as a product of two brackets.

Record your answer in your workbook.

The Big Idea

+5 XP

A quadratic equation $x^2 + bx + c = 0$ is solved by FACTORING into $(x - p)(x - q) = 0$, then applying the null factor law: if a product is zero, one of the factors must be zero. The solutions $x = p$ and $x = q$ are the roots — and they are exactly the $x$-intercepts of the parabola $y = x^2 + bx + c$.

The red parabola $y = x^2 - 5x + 6$ crosses the $x$-axis at $x = 2$ and $x = 3$. These are precisely the roots of $x^2 - 5x + 6 = 0$: factor to $(x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$. Algebra finds them; geometry shows them.

Roots of $x^2 - 5x + 6 = 0$ $=$ $x$-intercepts of $y = x^2 - 5x + 6$.

Null factor law

If $AB = 0$, then $A = 0$ or $B = 0$.

Roots = intercepts

Solving $= 0$ finds where the curve hits the $x$-axis.

Always check

Substitute each solution back — it must give $0$.

What You'll Master

objectives

Know

The null factor law: $AB = 0 \Rightarrow A = 0$ or $B = 0$
How to factor $x^2 + bx + c$ by finding two numbers that multiply to $c$ and add to $b$
Difference of two squares: $x^2 - a^2 = (x - a)(x + a)$

Understand

Why factoring + null factor law gives the roots
Why the roots of $y = 0$ are the $x$-intercepts of the parabola
Why a quadratic has at most TWO real solutions

Can Do

Solve $x^2 + bx + c = 0$ by factoring (trinomials and DOTS)
State the $x$-intercepts of a parabola from its quadratic equation
Verify solutions by substitution

Words You Need

vocabulary

Quadratic equationAn equation of the form $ax^2 + bx + c = 0$ with $a \neq 0$. Today we focus on $a = 1$.

Root / solutionA value of $x$ that makes the equation true. Quadratics have at most two real roots.

FactorWrite an expression as a product, e.g. $x^2 - 5x + 6 = (x - 2)(x - 3)$.

Null factor lawIf $AB = 0$, then $A = 0$ or $B = 0$ (or both).

TrinomialA three-term expression, e.g. $x^2 + 5x + 6$.

DOTSDifference of two squares: $a^2 - b^2 = (a - b)(a + b)$.

Spot the Trap

heads-up

Wrong: Solving $(x - 2)(x - 3) = 6$ by setting $x - 2 = 6$ and $x - 3 = 6$.

Right: The null factor law ONLY works when the product equals ZERO. Expand first, move everything to one side: $x^2 - 5x + 6 - 6 = 0 \Rightarrow x^2 - 5x = 0 \Rightarrow x(x - 5) = 0$.

Wrong: Factoring $x^2 - 9$ as $(x - 3)^2$.

Right: $x^2 - 9$ is a DIFFERENCE of squares: $(x - 3)(x + 3)$. $(x - 3)^2$ expands to $x^2 - 6x + 9$ — different.

Factoring Trinomials $x^2 + bx + c$

+5 XP

To factor $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$. Then $x^2 + bx + c = (x + p)(x + q)$ where $p + q = b$ and $pq = c$.

Example: $x^2 - 5x + 6$. Need two numbers with product $+6$, sum $-5$. Both negative (product positive, sum negative). Try $-2$ and $-3$: $(-2)(-3) = 6$, $(-2) + (-3) = -5$. So $x^2 - 5x + 6 = (x - 2)(x - 3)$.

To solve $x^2 - 5x + 6 = 0$: $(x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $x = 3$.

Find $p, q$ with $pq = c$, $p + q = b$. Then $(x + p)(x + q) = 0$.

Sign of $c$

$c > 0$: same signs. $c < 0$: opposite signs.

List factor pairs

Write all pairs of $c$, then pick the one that sums to $b$.

Expand to check

$(x - 2)(x - 3) = x^2 - 5x + 6$. Confirms the factoring.

Difference of Two Squares (DOTS)

+5 XP

When the quadratic has NO middle term and is a difference: $x^2 - a^2 = (x - a)(x + a)$.

Example: $x^2 - 9 = 0$. Recognise $9 = 3^2$, so $x^2 - 9 = (x - 3)(x + 3)$. Solve: $(x - 3)(x + 3) = 0 \Rightarrow x = 3$ or $x = -3$.

Geometrically, $y = x^2 - 9$ is the standard parabola shifted down 9 units — its $x$-intercepts are $\pm 3$.

Tip: DOTS only works for a difference. $x^2 + 9 = 0$ has no real solutions ($x^2 = -9$ has no real $x$).

$x^2 - a^2 = (x - a)(x + a)$. Roots: $x = \pm a$.

Spot the squares

$4, 9, 16, 25, 36, 49, 64, 81, 100$ — learn them on sight.

Always two roots

DOTS quadratics always give symmetric roots $\pm a$.

Sum ≠ DOTS

$x^2 + a^2$ does NOT factor over the reals.

Watch Me Solve It · 3 examples

Watch Me Solve It · Solve $x^2 - 5x + 6 = 0$

+15 XP per step

PROBLEM

Solve $x^2 - 5x + 6 = 0$ by factoring. State the $x$-intercepts of $y = x^2 - 5x + 6$.

1
Find two numbers
Need $pq = 6$ and $p + q = -5$. Both negative. Try $-2, -3$: $(-2)(-3) = 6$, $-2 + (-3) = -5$. ✓
2
Write factored form
$x^2 - 5x + 6 = (x - 2)(x - 3)$. So $(x - 2)(x - 3) = 0$.
3
Null factor law & check
$x - 2 = 0$ or $x - 3 = 0$, so $x = 2$ or $x = 3$. Check $x = 2$: $4 - 10 + 6 = 0$ ✓.
$x$-intercepts of the parabola: $(2, 0)$ and $(3, 0)$.

Answer$x = 2$ or $x = 3$. $x$-intercepts $(2, 0)$ and $(3, 0)$.

Watch Me Solve It · Difference of squares

+15 XP per step

PROBLEM

Solve $x^2 - 16 = 0$ by factoring using the difference of two squares.

1
Recognise DOTS
$16 = 4^2$. So $x^2 - 16 = x^2 - 4^2 = (x - 4)(x + 4)$.
2
Apply null factor law
$(x - 4)(x + 4) = 0 \Rightarrow x - 4 = 0$ or $x + 4 = 0$. So $x = 4$ or $x = -4$.
3
Check
$x = 4$: $16 - 16 = 0$ ✓. $x = -4$: $16 - 16 = 0$ ✓.
$y = x^2 - 16$ crosses the $x$-axis at $(\pm 4, 0)$.

Answer$x = \pm 4$.

Watch Me Solve It · Negative $c$

+15 XP per step

PROBLEM

Solve $x^2 + 2x - 15 = 0$.

1
Find two numbers
$pq = -15$, $p + q = 2$. Opposite signs (product negative). Try $5, -3$: $5 \times (-3) = -15$, $5 + (-3) = 2$. ✓
2
Factor and apply law
$(x + 5)(x - 3) = 0 \Rightarrow x + 5 = 0$ or $x - 3 = 0$.
3
Solve and check
$x = -5$ or $x = 3$. Check $x = -5$: $25 - 10 - 15 = 0$ ✓.
$x$-intercepts of $y = x^2 + 2x - 15$: $(-5, 0)$ and $(3, 0)$.

Answer$x = -5$ or $x = 3$.

Common Pitfalls

heads-up

Applying null factor law to non-zero

Writing $(x - 2)(x - 3) = 6$ then setting each factor to $6$. Wrong — the product is not $0$.

Fix: Always rearrange to $= 0$ first. Expand, collect, then factor.

Sign errors in factor pairs

For $x^2 - 5x + 6$, picking $+2, +3$ (which give $+5$, not $-5$).

Fix: Read the sign of $b$ carefully. If $b < 0$ and $c > 0$, BOTH factors are negative.

Confusing DOTS with perfect square

Writing $x^2 - 9 = (x - 3)^2$. Expanding gives $x^2 - 6x + 9 \neq x^2 - 9$.

Fix: A perfect square has THREE terms. A difference of squares has TWO and uses $(x - a)(x + a)$.

Missing one solution

From $x^2 = 16$, writing only $x = 4$.

Fix: Take BOTH square roots: $x = \pm 4$. A quadratic typically has TWO roots.

Copy Into Your Books

Null factor law

$AB = 0 \Rightarrow A = 0$ or $B = 0$
Only works when product $= 0$
Rearrange first if needed

Factor $x^2 + bx + c$

Find $p, q$ with $pq = c$
And $p + q = b$
Write $(x + p)(x + q)$

DOTS

$x^2 - a^2 = (x - a)(x + a)$
Roots: $x = \pm a$
Sum of squares: no real roots

Roots = intercepts

Solutions of $ax^2 + bx + c = 0$
are $x$-intercepts of $y = ax^2 + bx + c$
Always substitute to check

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Solving Quadratics

4 problems

Four quick problems: factoring, DOTS, and reading $x$-intercepts.

1 Solve $x^2 + 7x + 12 = 0$.
$pq = 12$, $p + q = 7$: pick $3, 4$. $(x + 3)(x + 4) = 0$.$x = -3$ or $x = -4$
2 Solve $x^2 - 25 = 0$.
DOTS: $(x - 5)(x + 5) = 0$.$x = \pm 5$
3 Solve $x^2 - x - 6 = 0$ and state the $x$-intercepts of $y = x^2 - x - 6$.
$pq = -6$, $p + q = -1$: pick $-3, 2$. $(x - 3)(x + 2) = 0$.$x = 3$ or $x = -2$; $x$-ints $(3, 0)$ and $(-2, 0)$
4 Solve $x^2 - 8x = 0$ (hint: common factor of $x$).
$x(x - 8) = 0$.$x = 0$ or $x = 8$

Complete in your workbook.

Quick Check · 5 questions

The solutions of $x^2 - 5x + 6 = 0$ are:

+10 XP

$x^2 - 9 = 0$ factored using the difference of two squares gives:

+10 XP

The solutions of $x^2 + 3x - 10 = 0$ are:

+10 XP

If $x = 4$ and $x = -1$ are the solutions of $x^2 - 3x - 4 = 0$, then the $x$-intercepts of $y = x^2 - 3x - 4$ are:

+10 XP

The solutions of $x^2 - 6x = 0$ are:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Solve $x^2 + 4x - 21 = 0$ by factoring. Show the factor pair and verify ONE solution by substitution.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. (a) Factor $x^2 - 49$ using DOTS. (b) Hence solve $x^2 - 49 = 0$. (c) State the $x$-intercepts of $y = x^2 - 49$.

Answer in your workbook.

ReasonHard3 MARKS

Q8. The parabola $y = x^2 + bx + c$ crosses the $x$-axis at $x = -2$ and $x = 7$. (a) Write the factored form of the quadratic. (b) Expand to find $b$ and $c$. (c) State the axis of symmetry.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $(x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $3$.

2. C — $(x - 3)(x + 3) = 0 \Rightarrow x = \pm 3$.

3. A — $(x + 5)(x - 2) = 0 \Rightarrow x = -5$ or $2$.

4. D — roots are the $x$-coordinates of the intercepts.

5. C — $x(x - 6) = 0 \Rightarrow x = 0$ or $x = 6$.

Show Your Working Model Answers

Q6 (3 marks): $pq = -21$, $p + q = 4$: choose $7$ and $-3$ [1]. Factor: $(x + 7)(x - 3) = 0$, so $x = -7$ or $x = 3$ [1]. Check $x = 3$: $9 + 12 - 21 = 0$ ✓ [1].

Q7 (3 marks): (a) $49 = 7^2$, so $x^2 - 49 = (x - 7)(x + 7)$ [1]. (b) $(x - 7)(x + 7) = 0 \Rightarrow x = 7$ or $x = -7$ [1]. (c) $x$-intercepts: $(7, 0)$ and $(-7, 0)$ [1].

Q8 (3 marks): (a) Roots $-2$ and $7$, so $y = (x + 2)(x - 7)$ [1]. (b) Expand: $(x + 2)(x - 7) = x^2 - 7x + 2x - 14 = x^2 - 5x - 14$, so $b = -5$, $c = -14$ [1]. (c) Axis: midpoint of $-2$ and $7$: $\tfrac{-2 + 7}{2} = \tfrac{5}{2} = 2.5$. Axis $x = 2.5$ [1].

Stretch Challenge · +25 XP, +10 coins

From Roots Back to the Equation

A parabola has $x$-intercepts at $x = -4$ and $x = 1$, and passes through the point $(0, -8)$. (a) Write a factored form $y = a(x - p)(x - q)$ with $p, q$ being the roots. (b) Substitute $(0, -8)$ to find $a$. (c) Expand to write the equation in $y = ax^2 + bx + c$ form. (d) Verify by solving your quadratic $= 0$ and checking the roots match.

Reveal solution

(a) Roots $-4, 1$: $y = a(x + 4)(x - 1)$. (b) Sub $(0, -8)$: $-8 = a(4)(-1) = -4a$, so $a = 2$. (c) $y = 2(x + 4)(x - 1) = 2(x^2 + 3x - 4) = 2x^2 + 6x - 8$. (d) Solve $2x^2 + 6x - 8 = 0 \Rightarrow x^2 + 3x - 4 = 0 \Rightarrow (x + 4)(x - 1) = 0 \Rightarrow x = -4$ or $x = 1$. Roots match.

Quick Review

Step 1

Rearrange to $= 0$

Step 2

Factor (trinomial or DOTS)

Step 3

Null factor law

Step 4

State both roots

Step 5

Substitute to check

Geometry

Roots $=$ $x$-intercepts of parabola

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