Lesson 18 ~25 min Unit 2 · Non-Linear +85 XP

Introduction to Quadratic Equations

From parabola to quadratic equation. The $x$-intercepts of $y = ax^2 + bx + c$ ARE the solutions of $ax^2 + bx + c = 0$. Solve simple quadratics by inspection and by factoring.

Today's hook: The parabola $y = x^2 - 9$ crosses the $x$-axis at $x = -3$ and $x = 3$. Notice anything? Those are exactly the SOLUTIONS of $x^2 - 9 = 0$. Coincidence?

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Think First

warm-up

A quadratic equation has the form $ax^2 + bx + c = 0$ (with $a \neq 0$). Its solutions are the values of $x$ that make the equation TRUE. Looking at the parabola $y = x^2 - 4$, where does it cross the $x$-axis? Now write down the solutions of $x^2 - 4 = 0$. What's the connection?

Record your answer in your workbook.

The Big Idea

+5 XP

The $x$-intercepts of the parabola $y = ax^2 + bx + c$ are exactly the SOLUTIONS (roots) of the equation $ax^2 + bx + c = 0$. The graph and the equation are two views of the same thing.

To find where $y = 0$ on a parabola, you set $ax^2 + bx + c = 0$ — that's a quadratic equation. The solutions ($x$-values) are called roots. Methods today: inspection (e.g. $x^2 = 9 \Rightarrow x = \pm 3$) and factoring (e.g. $x^2 - 5x + 6 = (x - 2)(x - 3) = 0$).

$x$-intercept on the graph $\Leftrightarrow$ solution of the equation $= 0$.

Roots = $x$-ints

Same numbers, two names.

$0$, $1$ or $2$ roots

Match the number of $x$-intercepts.

Two routes

Inspection (no $bx$ term) or factoring.

What You'll Master

objectives

Know

The general form $ax^2 + bx + c = 0$ ($a \neq 0$)
That $x$-intercepts of the parabola = roots of the equation
The null factor law: if $PQ = 0$ then $P = 0$ or $Q = 0$

Understand

Why $y = 0$ on the parabola corresponds to "equation $= 0$"
Why a quadratic can have $0$, $1$ or $2$ real solutions
Why factoring works thanks to the null factor law

Can Do

Solve $x^2 = k$ by inspection ($x = \pm \sqrt{k}$)
Solve $x^2 + bx + c = 0$ by factoring trinomials
Read roots off a parabola sketch

Words You Need

vocabulary

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

Root / solutionA value of $x$ that makes the equation true.

$x$-interceptWhere the curve crosses the $x$-axis; same numbers as the roots.

InspectionSolving by recognising a perfect square or simple form (e.g. $x^2 = 25$).

FactoringRewriting $x^2 + bx + c$ as $(x - r_1)(x - r_2)$ to reveal the roots.

Null factor lawIf $PQ = 0$, then $P = 0$ or $Q = 0$ — the basis of factoring.

Spot the Trap

heads-up

Wrong: $x^2 = 9 \Rightarrow x = 3$.

Right: $x^2 = 9 \Rightarrow x = \pm 3$. Both $3$ and $-3$ squared give $9$. ALWAYS take both signs.

Wrong: $(x - 2)(x - 3) = 0 \Rightarrow x - 2 = 0$ AND $x - 3 = 0$, so $x = 2$ AND $x = 3$ at the same time.

Right: The null factor law gives $x = 2$ OR $x = 3$ — two SEPARATE solutions to the equation.

Solving by Inspection

+5 XP

Inspection works when there's no $bx$ term — just $x^2 = $ number.

Move the constant to the other side: $x^2 = k$.
Take the square root of both sides — remember $\pm$.
Write both solutions: $x = \sqrt{k}$ or $x = -\sqrt{k}$.

If $k < 0$ (like $x^2 = -4$), there are NO real solutions. If $k = 0$, there's ONE solution: $x = 0$.

$x^2 - k = 0 \Rightarrow x^2 = k \Rightarrow x = \pm \sqrt{k}$.

Always $\pm$

Two solutions when $k > 0$.

$k < 0$: no real roots

No real number squares to a negative.

$k = 0$: one root

$x^2 = 0 \Rightarrow x = 0$ only.

Solving by Factoring

+5 XP

When there's a $bx$ term too, factor the trinomial $x^2 + bx + c$ into $(x - r_1)(x - r_2)$, then use the null factor law.

Move everything to one side: $x^2 + bx + c = 0$.
Find two numbers that MULTIPLY to $c$ and ADD to $b$.
Write the factored form: $(x + p)(x + q) = 0$.
Set each factor to $0$: $x = -p$ or $x = -q$.

Example: $x^2 - 5x + 6 = 0$. Two numbers multiplying to $6$, adding to $-5$: $-2$ and $-3$. So $(x - 2)(x - 3) = 0 \Rightarrow x = 2$ or $x = 3$.

$(x + p)(x + q) = 0 \Rightarrow x = -p$ or $x = -q$.

Product = $c$

Two numbers' product gives the constant.

Sum = $b$

Their sum gives the middle coefficient.

Null factor law

$PQ = 0 \Rightarrow P = 0$ or $Q = 0$.

Watch Me Solve It · 3 examples

Watch Me Solve It · Inspection

+15 XP per step

PROBLEM

Solve $x^2 - 9 = 0$ by inspection. Check using the parabola $y = x^2 - 9$.

1
Isolate $x^2$
$x^2 - 9 = 0 \Rightarrow x^2 = 9$.
2
Take $\pm \sqrt{\;}$
$x = \pm \sqrt{9} = \pm 3$. Two solutions: $x = 3$ and $x = -3$.
3
Graphical check
$y = x^2 - 9$ is a parabola with vertex $(0, -9)$, $x$-ints at $\pm 3$ — matches the roots.
Two routes, same answer: equation roots = parabola $x$-intercepts.

Answer$x = \pm 3$.

Watch Me Solve It · Factoring

+15 XP per step

PROBLEM

Solve $x^2 - 5x + 6 = 0$ by factoring.

1
Find the two numbers
Need product $= 6$, sum $= -5$. Try $-2$ and $-3$: $(-2)(-3) = 6$, $(-2) + (-3) = -5$. Both match.
2
Write factored form
$x^2 - 5x + 6 = (x - 2)(x - 3) = 0$.
3
Apply null factor law
$x - 2 = 0$ or $x - 3 = 0$, so $x = 2$ or $x = 3$.
Both check: $2^2 - 5(2) + 6 = 4 - 10 + 6 = 0$ \checkmark; $3^2 - 5(3) + 6 = 9 - 15 + 6 = 0$ \checkmark.

Answer$x = 2$ or $x = 3$.

Watch Me Solve It · Mixed signs

+15 XP per step

PROBLEM

Solve $x^2 + 2x - 8 = 0$ by factoring, then read off the $x$-intercepts of $y = x^2 + 2x - 8$.

1
Find the two numbers
Product $= -8$, sum $= 2$. Try $4$ and $-2$: $(4)(-2) = -8$ \checkmark, $4 + (-2) = 2$ \checkmark.
2
Factor and solve
$(x + 4)(x - 2) = 0 \Rightarrow x = -4$ or $x = 2$.
3
Read off $x$-intercepts
The parabola $y = x^2 + 2x - 8$ has $x$-intercepts at $(-4, 0)$ and $(2, 0)$ — same numbers as the roots.
Solving the equation IS finding the $x$-intercepts.

Answer$x = -4$ or $x = 2$; $x$-ints $(-4, 0)$ and $(2, 0)$.

Common Pitfalls

heads-up

Missing the $\pm$

$x^2 = 16 \Rightarrow x = 4$ — forgetting $x = -4$ too.

Fix: square root ALWAYS gives both signs when you solve $x^2 = k$ with $k > 0$.

Wrong sign on the root

$(x - 3) = 0 \Rightarrow x = -3$ — flipping the sign.

Fix: $(x - 3) = 0$ means $x = 3$. Add $3$ to both sides. The root has the OPPOSITE sign of the constant inside.

Forgetting to set $= 0$ first

Trying to factor $x^2 - 5x = 6$ directly without moving everything to one side.

Fix: rearrange to $x^2 - 5x - 6 = 0$ first, then factor. Null factor law needs zero on one side.

Copy Into Your Books

General form

$ax^2 + bx + c = 0$
$a \neq 0$
Up to $2$ real solutions

Inspection

$x^2 = k \Rightarrow x = \pm \sqrt{k}$
$k > 0$: $2$ roots
$k = 0$: $1$ root
$k < 0$: no real roots

Factoring

Product $= c$, sum $= b$
$(x + p)(x + q) = 0$
Null factor: $x = -p$ or $x = -q$

Graph connection

Roots = $x$-intercepts
$y = 0$ on the parabola
$0$, $1$ or $2$ depending on parabola

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Solve It

4 problems

Four quick problems on inspection and factoring.

1 Solve $x^2 = 25$.
$x = \pm \sqrt{25}$.$x = 5$ or $x = -5$
2 Solve $x^2 - 16 = 0$.
$x^2 = 16 \Rightarrow x = \pm 4$.$x = \pm 4$
3 Solve $x^2 - 7x + 12 = 0$ by factoring.
Product $= 12$, sum $= -7$: $-3$ and $-4$.$(x - 3)(x - 4) = 0 \Rightarrow x = 3$ or $x = 4$
4 Solve $x^2 + 3x - 10 = 0$ by factoring.
Product $= -10$, sum $= 3$: $5$ and $-2$.$(x + 5)(x - 2) = 0 \Rightarrow x = -5$ or $x = 2$

Complete in your workbook.

Quick Check · 5 questions

The solutions of $x^2 - 49 = 0$ are:

+10 XP

The solutions of $(x - 2)(x - 3) = 0$ are:

+10 XP

$x^2 - 5x + 6$ factors as:

+10 XP

The $x$-intercepts of $y = x^2 + bx + c$ are:

+10 XP

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

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyEasy3 MARKS

Q6. Solve each by inspection: (a) $x^2 = 36$, (b) $x^2 - 81 = 0$, (c) $x^2 + 4 = 0$ (explain).

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Solve $x^2 - 5x + 6 = 0$ by factoring. Then state the $x$-intercepts of the parabola $y = x^2 - 5x + 6$.

Answer in your workbook.

ReasonHard3 MARKS

Q8. A rectangle has length $(x + 5)$ cm and width $(x - 2)$ cm. (a) Write an expression for its area. (b) If the area is $0$, write the corresponding quadratic equation and solve it. (c) Which solution is physically meaningful for a real rectangle? Explain.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. B — $x = \pm 7$.

2. C — $x = 2$ or $x = 3$.

3. A — $(x - 2)(x - 3)$.

4. D — $x$-intercepts are the roots of the equation $= 0$.

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

Show Your Working Model Answers

Q6 (3 marks): (a) $x = \pm 6$ [1]. (b) $x^2 = 81 \Rightarrow x = \pm 9$ [1]. (c) $x^2 = -4$ has no real solutions: no real number squared gives a negative result [1].

Q7 (3 marks): Product $= 6$, sum $= -5$: $-2$ and $-3$. So $(x - 2)(x - 3) = 0$ [1]. Roots: $x = 2$ or $x = 3$ [1]. $x$-intercepts of the parabola: $(2, 0)$ and $(3, 0)$ — same numbers [1].

Q8 (3 marks): (a) Area $= (x + 5)(x - 2) = x^2 + 3x - 10$ cm$^2$ [1]. (b) $(x + 5)(x - 2) = 0 \Rightarrow x = -5$ or $x = 2$ [1]. (c) A rectangle needs positive length AND width: $x + 5 > 0$ ($x > -5$) and $x - 2 > 0$ ($x > 2$). Both solutions $x = -5$ and $x = 2$ give a degenerate rectangle (zero area) — the actual answer is that $x$ must be GREATER than $2$ for a real, nonzero rectangle. Neither root represents a real rectangle [1].

Stretch Challenge · +25 XP, +10 coins

From Sketch to Equation

A parabola has $x$-intercepts $(-3, 0)$ and $(4, 0)$, opens upward and has leading coefficient $a = 1$. (a) Write the parabola in factored form $y = (x - r_1)(x - r_2)$. (b) Expand to get $y = x^2 + bx + c$. (c) Hence write the QUADRATIC EQUATION whose roots are $-3$ and $4$. (d) Verify by substituting each root.

Reveal solution

(a) Roots $-3$ and $4$, so factors $(x + 3)$ and $(x - 4)$: $y = (x + 3)(x - 4)$. (b) Expand: $y = x^2 - 4x + 3x - 12 = x^2 - x - 12$. (c) Quadratic equation: $x^2 - x - 12 = 0$. (d) Sub $x = -3$: $9 - (-3) - 12 = 9 + 3 - 12 = 0$ \checkmark. Sub $x = 4$: $16 - 4 - 12 = 0$ \checkmark. Both check.

Quick Review

Quadratic

$ax^2 + bx + c = 0$, $a \neq 0$

Roots

$x$-intercepts on the parabola

Inspection

$x^2 = k \Rightarrow x = \pm \sqrt{k}$

Factoring

Product $c$, sum $b$

Null factor

$PQ = 0 \Rightarrow P = 0$ or $Q = 0$

Don't forget

$\pm$ on square roots

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