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Encyclopedia>Algebra>Basic Algebra>Algebraic Identities>What Is the Formula for Completing the Square?

Completing the square means adding two extra terms so you can write quadratic expressions in another way. The purpose of this is to make the expression simpler and easier to analyze. Here is how you remember the method:

Rule

Mnemonic Device for Completing the Square

1.: Halve it
2.: Square it
3.: Add it
4.: Subtract it

Note! If you add and subtract the same number, you don’t really change the value of the expression!

A quadratic expression is in the form

a x^{2} + b x + c

You make the expression involving $x$ ’s into a square when you write it in this form:

a {(x - x_{0})}^{2} + d,

so that

a x^{2} + b x + c = a {(x - x_{0})}^{2} + d .

When $d = 0$ , the expression is a perfect square, which is an expression where you can use the first or second algebraic identity of quadratic expressions. So you’re using these algebraic identities, only backwards. Here is how you do it, and the final formula:

Rule

Completing the Square

Say you have an expression on the form $a x^{2} + b x + c$ and are supposed to complete the square. First you factor out the coefficient $a$ and write the expression in the form $a (x^{2} + \frac{b}{a} x + \frac{c}{a})$ . This step can be ignored in the case where $a = 1$ .

Halve it:: $\frac{b}{a}$ is the number in front of the $x$ -term inside the parentheses. Divide this by 2. Then you get $\frac{b}{2 a}$ .
Square it:: $\frac{b}{2 a}$ is going to be squared. You then get ${(\frac{b}{2 a})}^{2}$ .
Add it:: Take this expression, ${(\frac{b}{2 a})}^{2}$ , and add it after the $\frac{b}{a} x$ -term.
Subtract it:: Take the same expression, ${(\frac{b}{2 a})}^{2}$ , and subtract it after $+ {(\frac{b}{2 a})}^{2}$ .

The whole expression looks like this:

\begin{array}{l} a (x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2} - {(\frac{b}{2 a})}^{2} + \frac{c}{a}) \\ = a {(x - x_{0})}^{2} + d . \end{array}

\begin{array}{l} a (x^{2} + \frac{b}{a} x + {(\frac{b}{2 a})}^{2} - {(\frac{b}{2 a})}^{2} + \frac{c}{a}) = a {(x - x_{0})}^{2} + d . \end{array}

Note! In the formula $a {(x - x_{0})}^{2} + d$ , you have $x_{0} = - \frac{b}{2 a}$ and $d = - \frac{b^{2}}{4 a} + c$ .

Example 1

Complete the square $x^{2} + 4 x + 3$ and write it in the form $a {(x - x_{0})}^{2} + d$

\begin{array}{l} x^{2} + 4 x + 3 \\ = x^{2} + 4 x + {(\frac{4}{2})}^{2} - {(\frac{4}{2})}^{2} + 3 \\ = x^{2} + 4 x + 2^{2} - 2^{2} + 3 \\ = {(x + 2)}^{2} - 4 + 3 \\ = {(x + 2)}^{2} - 1 \end{array}

\begin{array}{l} x^{2} + 4 x + 3 & = x^{2} + 4 x + {(\frac{4}{2})}^{2} - {(\frac{4}{2})}^{2} + 3 \\ = x^{2} + 4 x + 2^{2} - 2^{2} + 3 \\ = {(x + 2)}^{2} - 4 + 3 \\ = {(x + 2)}^{2} - 1 \end{array}

Example 2

Show that $x^{2} - 12 x + 36$ is a perfect square

\begin{array}{l} x^{2} - 12 x + 36 \\ = x^{2} - 12 x + {(\frac{- 12}{2})}^{2} - {(\frac{- 12}{2})}^{2} + 36 \\ = x^{2} - 12 x + 6^{2} - 6^{2} + 36 \\ = {(x - 6)}^{2} - 36 + 36 \\ = {(x - 6)}^{2} \end{array}

\begin{array}{l} x^{2} - 12 x + 36 & = x^{2} - 12 x + {(\frac{- 12}{2})}^{2} - {(\frac{- 12}{2})}^{2} + 36 \\ = x^{2} - 12 x + 6^{2} - 6^{2} + 36 \\ = {(x - 6)}^{2} - 36 + 36 \\ = {(x - 6)}^{2} \end{array}

Since

d = 0

, the expression

x^{2} - 12 x + 36

a perfect square.

Example 3

Find the minimum or the maximum of the function $f (x) = x^{2} - 3 x + 12$ by completing the square

\begin{array}{l} f (x) & = x^{2} - 3 x + {(\frac{- 3}{2})}^{2} - {(\frac{- 3}{2})}^{2} + 12 \\ = {(x - \frac{3}{2})}^{2} - {(\frac{- 3}{2})}^{2} + 12 \\ = {(x - \frac{3}{2})}^{2} - \frac{9}{4} + 12 \\ = {(x - \frac{3}{2})}^{2} - \frac{9}{4} + \frac{48}{4} \\ = {(x - \frac{3}{2})}^{2} - \frac{9 + 48}{4} \\ = {(x - \frac{3}{2})}^{2} - \frac{57}{4} \end{array}

\begin{array}{l} f (x) & = x^{2} - 3 x + {(\frac{- 3}{2})}^{2} - {(\frac{- 3}{2})}^{2} + 12 \\ = {(x - \frac{3}{2})}^{2} - {(\frac{- 3}{2})}^{2} + 12 \\ = {(x - \frac{3}{2})}^{2} - \frac{9}{4} + 12 \\ = {(x - \frac{3}{2})}^{2} - \frac{9}{4} + \frac{48}{4} \\ = {(x - \frac{3}{2})}^{2} - \frac{9 + 48}{4} \\ = {(x - \frac{3}{2})}^{2} - \frac{57}{4} \end{array}

Since

a > 0

you know that this function has a minimum. The

x

-value then becomes

x - \frac{3}{2} = 0

which gives us

x = \frac{3}{2}

. The

y

-value becomes

d = - \frac{57}{4}

. Then the minimum of the function has the following coordinates:

(\frac{3}{2}, - \frac{57}{4})

.

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