Quadratic Equations

Hong Kong

Stage 4 - Stage 5

Lesson

Completing the square is the process of rewriting a quadratic expression from the expanded form $x^2+2bx+c$`x`2+2`b``x`+`c` into the square form $\left(x+b\right)^2+c-b^2$(`x`+`b`)2+`c`−`b`2. The name comes from how the method was first performed by ancient Greek mathematicians who were actually looking for numbers to complete a physical square.

There are two ways to think about the completing the square method: visually (using a square), and algebraically. We will see that we can use this method to find the solutions to the quadratic equation and to more easily graph the function.

Consider the following quadratics:

$x^2+4x-5$`x`2+4`x`−5

$x^2+2x-3$`x`2+2`x`−3

$x^2-8x-20$`x`2−8`x`−20

Let's imagine we want to factorise them using the complete the square method.

The steps involved are:

- Move the constant term to the end
- Create a box and complete the square
- Keep the expression balanced
- Write the factorisation

As this is a visual method, its best to learn by watching the process in action, so here are three examples.

We wish to solve $x^2+4x-5=0$`x`2+4`x`−5=0 by first completing the square.

Completing the square on $x^2+4x-5$`x`2+4`x`−5:

Then we solve this algebraically.

$x^2+4x-5$x2+4x−5 |
$=$= | $0$0 |

$\left(x+2\right)^2-9$(x+2)2−9 |
$=$= | $0$0 |

$\left(x+2\right)^2$(x+2)2 |
$=$= | $9$9 |

$x+2$x+2 |
$=$= | $\pm3$±3 |

Then by subtracting $2$2 from both sides of the equation we have $x=1$`x`=1 or $x=-5$`x`=−5.

We wish to solve $x^2+2x-3=0$`x`2+2`x`−3=0 by first completing the square.

Completing the square on $x^2+2x-3$`x`2+2`x`−3:

Then we solve this algebraically.

$x^2+2x-3$x2+2x−3 |
$=$= | $0$0 |

$\left(x+1\right)^2-4$(x+1)2−4 |
$=$= | $0$0 |

$\left(x+1\right)^2$(x+1)2 |
$=$= | $4$4 |

$x+1$x+1 |
$=$= | $\pm2$±2 |

And by subtracting $1$1 from both sides of the equation we have $x=1$`x`=1 or $x=-3$`x`=−3.

We wish to solve $x^2-8x-20=0$`x`2−8`x`−20=0 by first completing the square.

Completing the square on $x^2-8x-20$`x`2−8`x`−20:

Then we solve this algebraically.

$x^2-8x-20$x2−8x−20 |
$=$= | $0$0 |

$\left(x-4\right)^2-36$(x−4)2−36 |
$=$= | $0$0 |

$\left(x-4\right)^2$(x−4)2 |
$=$= | $36$36 |

$x-4$x−4 |
$=$= | $\pm6$±6 |

We can add 4 to both sides of the equation to get $x=10$`x`=10 or $x=-2$`x`=−2.

This interactive will help you to visualise the process. Watch this video for an explanation .

Let's imagine we want to factorise the same quadratics as before, but this time using the algebraic version of the complete the square method.

The steps involved are:

- Move the constant term to the end
- Add half the coefficient of $x$
`x`and square it - Keep the expression balanced
- Complete the square

Solve the following quadratic equation *by completing the square*:

$x^2+18x+32=0$`x`2+18`x`+32=0

Solve $x^2-6x-16=0$`x`2−6`x`−16=0 *by completing the square*:

Solve for $x$`x` by first completing the square.

$x^2-2x-32=0$`x`2−2`x`−32=0

Solve the following quadratic equation *by completing the square*:

$4x^2+11x+7=0$4`x`2+11`x`+7=0

Write all solutions on the same line, separated by commas.

Enter each line of work as an equation.