GCSE(H),

The expression (*a* + 3)^{2} is an example of a **perfect square** and expands to *a*^{2} + 6*a* + 9.

The reverse process is called **completing the square**. Given *b*^{2} + 8*b* + 22, create a term plus a constant value that can be squared. This is useful as it gives the turning point, or vertex, of the graph of the function.

The constant term inside the bracket is half the coefficient of the *b* term. The square term is therefore (*b* + 4). Expanding (*b* + 4)^{2} gives *b*^{2} + 8*b* + 16. A further 6 needs to be added to this expression to obtain the original expression:

*b*^{2} + 8*b* + 22 = (*b* + 4)^{2} + 6.

The turning point of the quadratic graph is at (-4, 6): the value of the *x*-coordinate is the value of *b* that makes the squared term zero; the value of the *y*-coordinate is the value of the constant.

1. Complete the square for 4*a*^{2} + 8*a* - 12.

Answer: (2*a* + 2)^{2} - 16

Take the square root of the *x*^{2} coefficient: √4 = 2. This is the new *x* multiplier;

For the integer; half the *x* coefficient: `frac(1)(2)` x 8 = 4; then divide by the new x multiplier: 4 ÷ 2 = 2.

Result is (2*a* + 2)^{2} = 4*a*^{2} + 8*a* + 4. Need to get the integer to -12, so subtract 16.

2. Complete the square for *b*^{2} - 10*b*.

Answer: (*b* - 5)^{2} - 25

The square root of the *b*^{2} coefficient is 1. This is the new *b* multiplier.

For the integer: half the *b* coefficient, = -5.

Therefore (*b* - 5)^{2} = *b*^{2} - 10*b* + 25.

Need to adjust the integer value by -25 to obtain *b*^{2} - 10*b*.

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