Quadratic equations – factorisation and the pq formula

What is a quadratic equation?

A quadratic equation can be written as ax² + bx + c = 0, where a is not zero. The highest power of x is two. For example, x² − 5x + 6 = 0 is a quadratic equation. By contrast, 5x + 6 = 0 is a linear equation.

Here we work with real numbers. The equation can therefore have two different solutions, a repeated root or no solutions. Start by collecting all terms on one side and simplifying. Then choose a method that suits the expression.

If the equation is already x² = 9, you can take the square root directly: x = 3 or x = −3. Both numbers give nine when multiplied by themselves.

Solve by factorising

In x² − 5x + 6 = 0, we look for two numbers whose sum is −5 and whose product is 6. The numbers are −2 and −3.

A product is zero if at least one factor is zero. This is called the zero-product method.

1. Write (x − 2)(x − 3) = 0.
2. Set the first factor to zero: x − 2 = 0 gives x = 2.
3. Set the second factor to zero: x − 3 = 0 gives x = 3.

Check in the original equation: 2² − 5 · 2 + 6 = 0 and 3² − 5 · 3 + 6 = 0. Both solutions are correct.

Need to revise how brackets work? Read the guide to squaring identities and factorisation.

When do you use the pq formula?

For equations of the form x² + px + q = 0, you can use the pq formula. If the coefficient of x² is not one, first divide the entire equation by that coefficient.

In x² − 5x + 6 = 0, p = −5 and q = 6. The formula gives x = 2.5 ± √(6.25 − 6) = 2.5 ± 0.5. So x = 3 or x = 2.

If you have 2x² − 10x + 12 = 0 instead, divide every term by two. You then get the same equation as above.

The expression under the square root determines the number of real solutions: positive gives two, zero gives a repeated root and negative gives no real solutions.

Practise quadratic equations with answers

Problem 1

Solve x² − 7x + 12 = 0.

Problem 2

Solve x² + 4x + 4 = 0.

Problem 3

Does x² + 1 = 0 have a real solution?