Quadratic equations

A quadratic equation contains terms up to x^2. There are many ways to solve quadratics. All quadratic equations can be written in the form ax^2 + bx + c = 0 where a, b and c are numbers ( a cannot be equal to 0, but b and c can be).

Here are some examples of quadratic equations in this form:

  • 2x^2 - 2x - 3 = 0. a = 2, b = -2 and c = -3
  • 2x(x + 3) = 0. a = 2, b = 6 and c = 0 (in this example, the bracket can be expanded to 2x^2 + 6x = 0)
  • (2x + 1)(x - 5) = 0. a = 2, b = -9 and c = -5 (this will expand to 2x^2 - 9x - 5 = 0)
  • x^2 + 2 = 4. a = 1, b = 0 and c = -2
  • 3x^2 = 48. a = 3, b = 0 and c = -48 (in this example c = -48, but has been rearranged to the other side of the equation)

3x^2 = 48 is an example of a quadratic equation that can be solved simply.

3x^2 / 3 = 48 / 3

Divide both sides by 3

Solving by factorising

If the product of two numbers is zero then one or both of the numbers must also be equal to zero. Therefore if ab = 0, then a = 0 and/or b = 0.

If (x + 1)(x + 2) = 0, then x + 1 = 0 or x + 2 = 0, or both. Factorising quadratics will also be used to solve these equations.

Expanding the brackets (x + 2)(x + 3) gives x^2 + 3x + 2x + 6, which simplifies to x^2 + 5x + 6. Factorising is the reverse process of expanding brackets, so factorising x^2 + 5x + 6 gives (x + 2)(x + 3).

Example

Solve x(x + 3) = 0.

The product of x and x + 3 is 0, so x = 0 or x + 3 = 0, or both.

\begin{array}{rcl} x + 3 & = & 0 \\ -3 && -3 \\ x & = & -3 \end{array}

x = 0 or x = -3.

Question

Solve (x + 1)(x - 5) = 0.

The product of x + 1 and x - 5 is 0, so one or both brackets must also be equal to 0.

\begin{array}{rcl} x + 1 & = & 0 \\ -1 && -1 \\ x & = & -1 \end{array}

\begin{array}{rcl} x - 5 & = & 0 \\ + 5 && + 5 \\ x & = & 5 \end{array}

x = -1 or x = 5

Question

Solve x^2 + 7x + 12 = 0.

Begin by factorising the quadratic.

The quadratic will be in the form (x + a)(x + b) = 0.

Find two numbers with a product of 12 and a sum of 7.

3 \times 4 = 12, and 3 + 4 = 7, so a and b are equal to 3 and 4. This gives:

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

The product of x + 3 and x + 4 is 0, so x + 3 = 0 or x + 4 = 0, or both.

\begin{array}{rcl} x + 3 & = & 0 \\ - 3 && - 3 \\ x & = & -3 \end{array}

\begin{array}{rcl} x + 4 & = & 0 \\ - 4 && - 4 \\ x & = & -4 \end{array}

x = -3 or x = -4