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\(3x^2 = 48\) 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 \(x + 2 = 0\), meaning \(x = -1\) or \(x = -2\)

So, to solve the quadratic equation \(x^2 + 5x + 6 = 0\):

• Factorise the quadratic to get \((x + 2)(x + 3) = 0\).
• Therefore, \(x + 2 = 0\) or \(x + 3 = 0\), meaning \(x = -2\) or \(x = -3\) (always give both answers).

Solve \(x(x + 3) = 0\).

The product of \(x\) and \(x + 3\) is 0, so \(x = 0\) or \(x + 3 = 0\)

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

\(x = 0\) or \(x = -3\).

Solve \(x^2 – 6x + 8 = 0\)

\(x^2 – 6x – 8\) factorises as \((x – 2)(x – 4)\).

So \((x – 2)(x – 4) = 0\)

Therefore \(x = 2\) or \(x = 4\).