1MA1 · A18AlgebraFoundation & Higher

Solving Quadratic Equations

A quadratic equation contains a squared unknown and can have two, one or no real solutions. The form of the equation tells you whether factorising, a graph, the quadratic formula or completing the square is the most efficient method.

65 min Video by Maths Genie Pearson Edexcel GCSE Maths (1MA1)

The big picture

Solving a quadratic means finding the inputs that make its value zero. Algebraically these values make a factor zero; graphically they are the $x$ -coordinates where the parabola meets the $x$ -axis. The two viewpoints describe the same roots.

Edexcel questions often hide the quadratic inside geometry or another equation. The dependable strategy is to form the equation, rearrange it into $a x^{2} + b x + c = 0$ , choose a suitable method and interpret both solutions in context.

Solving Quadratics by FactorisingWatch the walkthrough, then read the notes below.

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What you'll learn

Rearrange and solve quadratic equations by factorising
Use the zero-product rule and check both roots
Find approximate solutions using a graph
Use the quadratic formula and completing the square at Higher tier
Reject solutions that are impossible in context

Factorise, then use the zero-product rule

First rearrange until one side is zero. Factorise the quadratic, then set each factor equal to zero. This works because a product is zero only when at least one of its factors is zero.

Do not divide by a factor containing $x$ : that can silently discard one solution. From $x (x - 5) = 0$ , division by $x$ would lose the valid root $x = 0$ .

A B = 0 \Rightarrow A = 0 or B = 0

The zero-product rule.

1Rearrange:

x^{2} + 3 x - 10 = 0

2Factorise:

(x + 5) (x - 2) = 0

3Set each bracket to zero:

x + 5 = 0

x - 2 = 0

Answer

x = - 5

x = 2

Why this works

Factorising reveals the values that make the quadratic zero. Each linear factor contributes one possible root.

Harder factorising and contextual roots

At Higher tier, the coefficient of $x^{2}$ may be greater than 1. Factorise carefully, solve each linear bracket, and substitute the results into the original equation if a check is needed.

A mathematically valid root may be impossible in context. Length, time and population cannot usually be negative, so state why a root is rejected rather than simply omitting it.

1Form the equation:

x (2 x + 3) = 35

2Rearrange:

2 x^{2} + 3 x - 35 = 0

3Factorise:

(2 x - 7) (x + 5) = 0

4So

x = 3.5

x = - 5

. Reject

- 5

because a width cannot be negative.

5Length

= 2 (3.5) + 3 = 10

Answer

3.5

cm by

10

Tip — An answer in context needs the requested quantity and its unit—not merely the value of $x$ .

The quadratic formula (Higher)

For $a x^{2} + b x + c = 0$ , the quadratic formula works whether or not the expression factorises neatly. Identify $a$ , $b$ and $c$ only after arranging the equation in descending powers and making one side zero.

Substitute negative coefficients using brackets, and keep the entire numerator over $2 a$ . Use the $\pm$ to calculate both solutions, then round only at the end to the requested accuracy.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The quadratic formula.

a = 2

b = 3

c = - 4

2Substitute:

x = \frac{- 3 \pm 3 ^{2} - 4 ( 2 ) ( - 4 )}{2 ( 2 )}

3Simplify:

x = \frac{- 3 \pm 41}{4}

4Evaluate both branches and round at the end.

Answer

x = 0.851

x = - 2.35

(3 s.f.)

Why this works

The expression $b^{2} - 4 a c$ under the square root controls how many real roots exist: positive gives two, zero gives one repeated root, and negative gives no real roots.

Completing the square and graphical solutions (Higher)

Completing the square rewrites $x^{2} + b x + c$ as $(x + p)^{2} + q$ . It is especially useful when a question asks for an exact solution or connects the equation to the graph’s turning point.

A graph gives approximate roots where the curve crosses $y = 0$ . To solve two equations graphically, locate their intersection points and read the corresponding $x$ -coordinates with the accuracy the scale permits.

x^{2} + 6 x + 2 = (x + 3)^{2} - 9 + 2 = (x + 3)^{2} - 7

2Set equal to zero:

(x + 3)^{2} = 7

3Square-root both sides:

x + 3 = \pm 7

Answer

x = - 3 \pm 7

Think like an examiner

Always show the equation rearranged to zero and the factorised line; both may carry method marks.
Give both roots unless the context rules one out, and explicitly explain any rejection.
For formula questions, write the substitution before the decimal answers and state the requested accuracy.
A graphical solution is approximate; do not report more precision than the graph scale supports.

Remember these

A B = 0 \Rightarrow A = 0 or B = 0

Zero-product rule.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Quadratic formula (Higher).

x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}

Completing-square pattern (Higher).

Watch out for these

Factorising while the equation is not equal to zero.

Rearrange to

a x^{2} + b x + c = 0

before using the zero-product rule.

From

(x + 4) (x - 3) = 0

, writing

x = 4

x = - 3

Solve each bracket:

x = - 4

x = 3

Rounding intermediate formula values.

Keep the calculator display or exact surd until the final answer, then round once.

Stretch yourself

The equation $x^{2} + k x + 16 = 0$ has exactly one real solution. Find the possible values of $k$ .

Hint — Exactly one real solution means the expression under the square root in the quadratic formula equals zero.

Questions students ask

How do I choose a solving method?

Factorise when the factors are apparent; use a graph when an approximate solution is requested; use the formula when factorising is difficult; complete the square when an exact form or turning point is useful.

Why can a quadratic have fewer than two real roots?

A parabola may cross the

x

-axis twice, touch it once, or miss it entirely. These cases correspond to a positive, zero or negative discriminant.

Key takeaways

Rearrange to zero before factorising and applying the zero-product rule.
The quadratic formula solves any real quadratic when the discriminant is non-negative.
Completing the square gives exact roots and exposes the turning-point form.
Interpret every solution in its original context.

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