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To solve an equation of the form \(x^2 + bx + c = 0\) consider the expression \(\left(x + \frac{b}{2}\right)^2 + c\).

\(\left(x + \frac{b}{2}\right)^2 + c\) expands to \(x^2 + bx + \left(\frac{b}{2}\right)^2 + c\), which is the same as the left-hand side of the original equation but with an extra term \(\left(\frac{b}{2}\right)^2\).

So the equation \(ax^2 + bx + c = 0 \) becomes \(\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c = 0\).

This can be rearranged to give \(\left(x + \frac{b}{2}\right)^2 = \left(\frac{b}{2}\right)^2 - c\) which can then be solved by taking the square root of both sides.

Solve \(x^2 + 6x - 10 = 0\) by completing the square.

• Halving the value of \(b\) (the coefficient of \(x\)) gives \(6 \div 2 = 3\)
• Thus the completed square form starts with \((x + 3)^2\)
• But \((x + 3)^2\) gives \(x^2 + 6x + 9\) when we want \(x^2 + 6x – 10\).
• To correct this (completing the square) we need to subtract 19, giving \((x + 3)^2 – 19 = 0\).

Rearrange this quadratic to get \((x + 3)^2\) alone on the left-hand side by adding 19 to each side.

\(x = -3 \pm \sqrt{19}\)

The first solution is: \(x = -3 + \sqrt{19} = 1.36\) (2 dp)

The second solution is: \(x = -3 - \sqrt{19} = -7.36\) (2 dp)