Completing the Square - a MaS Tutor.

Completing the square is useful method for solving several problems and finding new information about a quadratic expression. In general, completing the square changes an expression with two terms involving $x$ into one where $x$ appears only once.

x^{2} + b x + c = {(x + p)}^{2} + q

$p$ and $q$ can be used to locate the vertex of the parabola, i.e., $(-p, q)$ . Alternatively, the completed expression can be used in an equation to find the solutions.

In order to convert the left-hand side to the right we need only concentrate on the first two terms, $x^{2} + b x$ , to start with.

Write down $(x +$ ;
then $b$ divided by two;
followed by $)^{2}$ ;
now subtract ${(\frac{b}{2})}^{2}$ .

Having completed this method we then have to substitute back into the original problem
and, perhaps, simplify. There is a proof of this method.

Example 1:

x^{2} + 4 x

= {(x + \frac{4}{2})}^{2} - {(\frac{4}{2})}^{2}

= {(x + 2)}^{2} - 4

Example 2:

x^{2} - 6 x + 10

= {(x - \frac{6}{2})}^{2} - {(\frac{6}{2})}^{2} + 10

= {(x - 3)}^{2} - 9 + 10

= {(x - 3)}^{2} + 1

Example 3:

2 x^{2} - 9 x - 1

= 2 (x^{2} - \frac{9}{2} x) - 1

= 2 ({(x - \frac{9}{4})}^{2} - {(\frac{9}{4})}^{2}) - 1

= 2 {(x - \frac{9}{4})}^{2} - 2 (\frac{81}{16}) - 1

= 2 {(x - \frac{9}{4})}^{2} - \frac{81}{8} - 1

= 2 {(x - \frac{9}{4})}^{2} - \frac{89}{8}

Completing the Square (CtS) Video: Practice Questions 1; Practice Questions 2; Textbook Exercises; Textbook Exercises - solving with CtS
Quadratic graph (CtS) Video
Solving Quadratics by completing the square Video:: Practice Questions;; Textbook Exercise.; (Care of corbettmaths.com)
Completing the Square Video:: Exam Questions;; Exam Questions Booklet;; Solutions.; (Care of www.mathsgenie.co.uk)

Completing the Square Proof

x^{2} + b x = {(x + p)}^{2} + q

= x^{2} + 2 p x + p^{2} + q

Juxtapose the first and last expressions ...

x^{2} + b x

x^{2} + 2 p x + p^{2} + q

\Rightarrow 2 p = b

& p^{2} + q = 0

∴ p = \frac{b}{2}

& q = - {(\frac{b}{2})}^{2}

In summary

x^{2} + b x = {(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2}

... this is completed square form.

Further confirmation can be achieved by multiplying out the right hand side of the above

{(x + \frac{b}{2})}^{2} - {(\frac{b}{2})}^{2}

= x^{2} + b x + \frac{b^{2}}{4} - \frac{b^{2}}{4}

= x^{2} + b x

... as required