Completing the square is a useful technique for solving quadratic equations. It is a more powerful technique than factorisation because it can be applied to equations that do not factorise.

When completing the square, an expression like,

ax^2 + bx + c is written in the form (Ax + B)^2 + C.

We will begin with the simple example where *a* = 1. In this case we will write expressions in the form

x^2 + bx + c as (x + B)^2 + C

If we expand (x + B)^2 + C we get x^2 + 2Bx + B^2 + C.

Comparing this with x^2 + bx + c shows that

b = 2B and c = B^2 + C

which gives *B* = and C = c - B^2

Using these two results we can now set about completing the square in some simple cases.

Write each of the following expressions in the form (x + *B*)^2 + *C*.

(a)

x^2 + 6x + 1

Comparing x^2 + 6x + 1 with x^2 + bx + c we see that *b* = 6 and *c* = 1 in this case, so

*B* = = = 3 and
C = c - B^2 = 1 - 3^2 = -8.

Therefore x^2 + 6x + 1 = (x + 3)^2 - 8.

(b)

x^2 + 4x - 2

Here *b* = 4 and *c* = –2, so

*B* = = = 2 and
C = c - B^2 = (-2) - 2^2 = -6.

Therefore x^2 + 4x - 2 = (x + 2)^2 - 6.

(c)

x^2 + 2x

Here *B* = = 1 and C = 0 - 1^2 = -1,

so x^2 + 2x = (x + 1)^2 - 1.

Solve the following equations by completing the square.

(a)

x^2 - 4x - 5 = 0

Completing the square gives,

x^2 - 4x - 5 = (x - 2)^2 - 9

Now we can solve the equation

(x – 2)^{2} | = 9 |

x – 2 | = ± |

x – 2 | = ± 3 |

x | = 2 ± 3 |

sox | = 5 or –1 |

(b)

x^2 + 6x - 1 = 0

Completing the square gives,

x^2 + 6x - 1 = (x + 3)^2 - 10

Now we can solve the equation

(x + 3)^{2} | = 10 |

x + 3 | = ± |

x | = –3 ± |

sox | = 0.162 or –6.162 to 3 decimal places |

Note: To type indeces on this page use ^ sign. e.g. *n*^{2}: