2.1PureCore

Solving Quadratic Equations

A quadratic equation has the form $a x^{2} + b x + c = 0$ . There are three standard routes to a solution — factorising, the quadratic formula, and completing the square. Knowing when to use each one is the key skill here.

35 min Video by Zeeshan Zamurred Quadratics

Edexcel AS Maths: 2.1 Solving Quadratic EquationsWatch the full walkthrough before the notes below.

Open on YouTube

What you'll be able to do

Recognise a quadratic equation in standard form
Solve quadratics by factorising
Solve quadratics using the quadratic formula
Solve disguised (hidden) quadratic equations

Standard form and the zero-product rule

First rearrange every quadratic so that one side equals zero: $a x^{2} + b x + c = 0$ . The whole method depends on this.

Once a quadratic is factorised into two brackets, the $zero-product rule$ applies: if two things multiply to give zero, at least one of them must be zero.

If (x - p) (x - q) = 0 then x = p or x = q .

Set each bracket equal to zero to read off the two roots.

Solving by factorising

If the quadratic factorises with whole numbers, this is the fastest method. Factorise, then set each bracket to zero.

1Two numbers multiplying to

- 15

and adding to

2

: that is

5

and

- 3

2Factorise:

(x + 5) (x - 3) = 0

3Set each bracket to zero:

x + 5 = 0

x - 3 = 0

Answer

x = - 5

x = 3

Tip — Always rearrange to “= 0” first. You cannot use the zero-product rule on, say, (x+5)(x−3) = 7.

The quadratic formula

When a quadratic will not factorise nicely, use the formula. It works for $every$ quadratic, so it is a reliable fallback.

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

Read off

a

b

and

c

from

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

and substitute carefully.

1Here

a = 2

b = 3

c = - 1

2Discriminant:

b^{2} - 4 a c = 9 - 4 (2) (- 1) = 17

3So

x = \frac{- 3 \pm 17}{4}

Answer

x = 0.28

x = - 1.78

Tip — Watch negative signs: −4ac with a negative c becomes a plus. This is the most common formula slip.

Hidden (disguised) quadratics

Some equations are quadratics in disguise — for example in $x^{4}$ or $x$ . Use a substitution to turn them into an ordinary quadratic, solve, then convert back.

1Let

y = x^{2}

, so the equation becomes

y^{2} - 5 y + 4 = 0

2Factorise:

(y - 1) (y - 4) = 0

, so

y = 1

y = 4

3Convert back:

x^{2} = 1 \Rightarrow x = \pm 1

;

x^{2} = 4 \Rightarrow x = \pm 2

Answer

x = \pm 1, \pm 2

Formula recap

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

Standard form — always rearrange to this.

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

The quadratic formula.

(x - p) (x - q) = 0 \Rightarrow x = p, q

Zero-product rule.

Common mistakes to avoid

Solving without rearranging to = 0 first.

Move every term to one side so the equation reads ... = 0 before factorising.

Forgetting the ± in the quadratic formula, giving only one root.

The ± gives two roots — always write both.

Key takeaways

Rearrange to ax² + bx + c = 0 first.
Try factorising; if it is awkward, use the quadratic formula.
The ± in the formula produces two roots.
Spot disguised quadratics and use a substitution.

Test yourself

Ready to lock in Solving Quadratic Equations? Pick a mode and earn XP & Dobloons.