������̳

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).

• \(2x^2 - 2x - 3 = 0\). \(a = 2\), \(b = -2\) 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.

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 \((x + 2)(x + 3)\).

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 \(x = -3\)