Solving Quadratic Equations

A quadratic equation contains a term in $x^{2}$x2 and can be written in the form $ax^{2} + bx + c = 0$ax2+bx+c=0. This lesson covers solving by factorising, the quadratic formula, and completing the square (Higher), along with interpreting solutions in context.

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Key Vocabulary

Term	Meaning
Quadratic equation	An equation of the form $ax^{2} + bx + c = 0$ax2+bx+c=0
Root / solution	A value of x that satisfies the equation
Discriminant	$b^{2} - 4ac$b2−4ac — determines the number of solutions
Completing the square	Writing $ax^{2} + bx + c$ax2+bx+c in the form $a(x + p)^{2} + q$a(x+p)2+q

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Solving by Factorising

If a product of factors equals zero, at least one factor must be zero.

Method

1. Rearrange so one side equals 0.
2. Factorise.
3. Set each factor equal to 0 and solve.

Worked Example 1

Solve: $x^{2} + 5x + 6 = 0$x2+5x+6=0

Step 1: Factorise: (x + 2)(x + 3) = 0

Step 2: $x + 2 = 0 \to x = -2$x+2=0→x=−2, or $x + 3 = 0 \to x = -3$x+3=0→x=−3

Answer: x = −2 or x = −3

Worked Example 2

Solve: $x^{2} - 7x = 0$x2−7x=0

Factorise: x(x − 7) = 0

x = 0 or x = 7

Answer: x = 0 or x = 7

(Don't divide both sides by x — you'd lose the solution x = 0.)

Worked Example 3

Solve: $2x^{2} + 5x - 3 = 0$2x2+5x−3=0

Factorise (using the grouping method): ac = −6. Numbers: 6 and −1.

$2x^{2} + 6x - x - 3 = 0$2x2+6x−x−3=0

2x(x + 3) − 1(x + 3) = 0

(2x − 1)(x + 3) = 0

x = 1/2 or x = −3

Answer: x = ½ or x = −3

Worked Example 4

Solve: $x^{2} = 6x - 9$x2=6x−9

Rearrange: $x^{2} - 6x + 9 = 0$x2−6x+9=0

$(x - 3)(x - 3) = 0 \to (x - 3)^{2} = 0$(x−3)(x−3)=0→(x−3)2=0

x = 3 (repeated root)

Answer: x = 3

Choosing a Method to Solve a Quadratic

flowchart TD
    EQ[Equation to solve] --> CHECK1{"Is it of the form<br/>ax squared + bx + c = 0?"}
    CHECK1 -->|No: rearrange first| REARR[Move all terms to one side]
    REARR --> CHECK1
    CHECK1 -->|Yes| LINQ{Highest power of x?}
    LINQ -->|Power 1: linear| LIN["Use inverse operations<br/>see Lesson 3"]
    LINQ -->|Power 2: quadratic| TRY{Does it factorise easily?}
    TRY -->|Yes| FAC["FACTORISE then set<br/>each factor to zero"]
    TRY -->|No / not obvious| FORM{Exact answer needed?}
    FORM -->|Yes, exact / surd form| CSQ["COMPLETE THE SQUARE H<br/>or quadratic formula"]
    FORM -->|Decimal allowed| QFORM["QUADRATIC FORMULA<br/>x = minus b plus or minus root b sq minus 4ac all over 2a"]
    FAC --> CHECKD{Discriminant b sq minus 4ac}
    QFORM --> CHECKD
    CSQ --> CHECKD
    CHECKD -->|Greater than 0| TWO[Two distinct real roots]
    CHECKD -->|Equal to 0| ONE[One repeated root]
    CHECKD -->|Less than 0| NONE[No real roots]

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The Quadratic Formula

When the equation cannot be factorised easily (or to confirm your factorisation), use the formula:

$x = (-b \pm \sqrt{b^{2} - 4ac}) / 2a$x=(−b±b2−4ac​)/2a

This formula is given on the Edexcel formulae sheet.

Worked Example 5

Solve: $x^{2} + 3x - 7 = 0$x2+3x−7=0. Give answers to 2 decimal places.

a = 1, b = 3, c = −7

Discriminant: $b^{2} - 4ac = 9 + 28 = 37$b2−4ac=9+28=37

$x = (-3 \pm \sqrt{37}) / 2$x=(−3±37​)/2

x = (−3 + 6.083) / 2 = 3.083 / 2 = 1.54 (2 d.p.)

x = (−3 − 6.083) / 2 = −9.083 / 2 = −4.54 (2 d.p.)

Worked Example 6

Solve: $2x^{2} - 5x + 1 = 0$2x2−5x+1=0. Give answers to 3 significant figures.

a = 2, b = −5, c = 1

Discriminant: 25 − 8 = 17

$x = (5 \pm \sqrt{17}) / 4$x=(5±17​)/4

x = (5 + 4.123) / 4 = 2.28 (3 s.f.)

x = (5 − 4.123) / 4 = 0.219 (3 s.f.)

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The Discriminant

The value $b^{2} - 4ac$b2−4ac tells you how many real solutions there are:

Value of $b^{2} - 4ac$b2−4ac	Number of solutions	What it means
> 0	2 distinct real solutions	The graph crosses the x-axis at two points
= 0	1 repeated real solution	The graph touches the x-axis at one point
< 0	No real solutions	The graph does not cross the x-axis

Worked Example 7

Show that $x^{2} + 2x + 5 = 0$x2+2x+5=0 has no real solutions.

$b^{2} - 4ac = 4 - 20 =$b2−4ac=4−20= −16 < 0

Since the discriminant is negative, there are no real solutions. ∎

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Completing the Square [H]

Write $x^{2} + bx + c$x2+bx+c in the form $(x + p)^{2} + q$(x+p)2+q, where p = b/2.

Method

1. Take half the coefficient of x: p = b/2.
2. Write $(x + p)^{2}$(x+p)2 — this expands to $x^{2} + 2px + p^{2}$x2+2px+p2.
3. Adjust: subtract $p^{2}$p2 and add c.

Worked Example 8

Write $x^{2} + 6x + 2$x2+6x+2 in the form $(x + p)^{2} + q$(x+p)2+q.

Half of 6 is 3, so: $(x + 3)^{2} = x^{2} + 6x + 9$(x+3)2=x2+6x+9

We need $x^{2} + 6x + 2 = (x + 3)^{2} - 9 + 2 =$x2+6x+2=(x+3)2−9+2= $(x + 3)^{2} - 7$(x+3)2−7

Worked Example 9

Solve $x^{2} + 6x + 2 = 0$x2+6x+2=0 by completing the square. Give exact answers.

From above: $(x + 3)^{2} - 7 = 0$(x+3)2−7=0

$(x + 3)^{2} = 7$(x+3)2=7

$x + 3 = \pm\sqrt{7}$x+3=±7​

$x = -3 \pm \sqrt{7}$x=−3±7​

Answer: $x = -3 + \sqrt{7}$x=−3+7​ or $x = -3 - \sqrt{7}$x=−3−7​

Worked Example 10 — When $a \neq 1 [H]$a=1[H]

Write $2x^{2} + 12x + 7$2x2+12x+7 in the form $a(x + p)^{2} + q$a(x+p)2+q.

Factor out the 2 from the x-terms: $2[x^{2} + 6x] + 7$2[x2+6x]+7

Complete the square inside: $2[(x + 3)^{2} - 9] + 7$2[(x+3)2−9]+7

$= 2(x + 3)^{2} - 18 + 7$=2(x+3)2−18+7

= $2(x + 3)^{2} - 11$2(x+3)2−11

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Interpreting Solutions in Context

Some problems give a quadratic in a real-world context. Always check whether both solutions are valid.

Worked Example 11

A ball is thrown upward. Its height h metres after t seconds is $h = 20t - 5t^{2}$h=20t−5t2.

Find when the ball hits the ground.

Set h = 0: $20t - 5t^{2} = 0$20t−5t2=0

5t(4 − t) = 0

t = 0 or t = 4

t = 0 is the start, so the ball hits the ground at t = 4 seconds.

Worked Example 12

The area of a rectangle is $60 cm^{2}$60cm2. Its length is 4 cm more than its width. Find the dimensions.

Let width = w. Then length = w + 4.

w(w + 4) = 60

$w^{2} + 4w - 60 = 0$w2+4w−60=0

(w + 10)(w − 6) = 0

w = −10 or w = 6

Width cannot be negative, so w = 6 cm, length = 10 cm.

Check: $6 \times 10 = 60$6×10=60 ✓

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Common Mistakes and Misconceptions