Quadratic equations

A quadratic equation contains only terms up to and including \(x^2\). 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.

  • \(2x^2 - 2x - 3 = 0. a = 2, b = -2\) and \(c = -3\)
  • \(2x(x+3) = 0\). This will expand to \(2x^2 + 6x = 0. a = 2, b = 6\) and \(c = 0\)
  • \((2x + 1)(x - 5) = 0\). This will expand to \(2x^2 - 9x - 5 = 0. a = 2, b = -9\) and \(c = -5\)
  • \(x^2 + 2 = 4\). This will rearrange to \(x^2 - 2 = 0. a = 1, b = 0\) and \(c = -2\)
  • \(3x^2 = 48\). This will rearrange to \(3x^2 – 48 = 0\). \(a = 3, b = 0\) 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 the equation.

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 \(x + 1 = 0\) or \(x – 5 = 0\).

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

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

\(x = -1\) 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 \(x = -4\)