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Quadratic expressions, functions, and equations can all be manipulated in useful ways with a method called completing the square. This method is based on the concept of *perfect square trinomials.*

Just as a perfect square integer, such as $9,$ can be written as the square of integer square roots, a **perfect square trinomial** is a trinomial which can be written as the square of a binomial.
$9↓(3)(3)↓3_{2} x_{2}+2bx+b_{2}↓(x+b)(x+b)↓(x+b)_{2} $

To be able to rewrite an expression to include a perfect square trinomial, it is first necessary to be able to recognize them. This can be achieved by expanding the square of a general binomial. The factoring is then done in the opposite direction. There are two kinds of perfect square trinomials, leading to different signs between the terms in the binomial.

If a trinomial is in the form $a_{2}+2ab+b_{2},$ where $a$ and $b$ are variables or positive numbers, it is a perfect square trinomial that can be factored as $(a+b)_{2}.$ This is shown by expanding the squared binomial.

$(a+b)_{2}$

SplitIntoFactorsSplit into factors

$(a+b)(a+b)$

MultParMultiply parentheses

$a⋅a+a⋅b+b⋅a+b⋅b$

MultiplyMultiply

$a_{2}+ab+ab+b_{2}$

SimpTermsSimplify terms

$a_{2}+2ab+b_{2}$

Thus, the trinomial and the square are equal.

If a trinomial instead is in the form $a_{2}−2ab+b_{2},$ it is also a perfect square trinomial and can be factored as $(a−b)_{2}.$ This is shown by expanding the squared binomial.

$(a−b)_{2}$

SplitIntoFactorsSplit into factors

$(a−b)(a−b)$

MultParMultiply parentheses

$a⋅a+a⋅(-b)+(-b)⋅a+(-b)⋅(-b)$

MultiplyMultiply

$a_{2}−ab−ab+b_{2}$

SimpTermsSimplify terms

$a_{2}−2ab+b_{2}$

The trinomial and the square are indeed equal.

The expressions in the table are perfect square trinomials. Complete the expressions to ensure equivalence between corresponding standard forms and factored forms.

standard form | factored form |
---|---|

$x_{2}−2x+1$ | $(x$__ ___$)_{2}$ |

$x_{2}−$___$x+$___ | $(x−7)_{2}$ |

$x_{2}−10x+$___ | $(x$__ ___$)_{2}$ |

$x_{2}+$___$x+100$ | $(x$__ ___$)_{2}$ |

Show Solution

Here the same expressions are written both in standard form and in factored form, but one or more terms are missing in each expression. To identify the missing term(s) the rules for factoring a perfect square trinomial are useful.

and

$a_{2}−2ab+b_{2}=(a−b)_{2}$

To help us identify the terms in the rule in the first example, we can for the expression $x_{2}−2x+1$ rewrite $1$ as $1_{2}$ and $2x$ as $2⋅x⋅1.$

$x_{2}−2⋅x⋅1+1_{2}=(x$__ ___$)_{2}$

To make the expressions match each other we will fill in $−1$ where the blanks are.

$x_{2}−2⋅x⋅1+1_{2}=(x−1)_{2}$

We can use this reasoning to complete the other perfect square trinomials in the table. Next, we'll consider the second row.

$x_{2}−$___$x+$___ $=(x−7)_{2}$

To fill in the second blank we use that $b$ must be $7,$ which gives us that $b_{2}=7_{2}.$ When we study the first blank we can match $a$ with $x.$ What is left is $2⋅b=2⋅7.$

$x_{2}−2⋅x⋅7+7_{2}=(x−7)_{2}$

This we prefer writing as $x_{2}−14x+49=(x−7)_{2}.$ Let's continue with the third row

$x_{2}−10x+$___$=(x$__ ___$)_{2}$

Here the terms $a$ and $x$ match each other. Since the terms $2ab$ and $10x$ must match each other we find that $2b=10,$ or $b=5.$ Let's use this to fill in the blanks in the expression.

$x_{2}−10x+5_{2}=(x−5)_{2}$

By squaring $5$ we get that the second row reads $x_{2}−10x+25=(x−5)_{2}.$ We are now going to deal with the last row in the table.

$x_{2}+$___$x+100$$=(x$__ ___$)_{2}$

The lines match each other when $b=10,$ giving us the first blank $2⋅10=20$ and the second blank $+10.$ Let's write them together with the blanks filled in.

$x_{2}+20x+100=(x+10)_{2}$

We'll summarize our results by adding the found values to the table.

standard form | factored form |
---|---|

$x_{2}−2x+1$ | $(x−1)_{2}$ |

$x_{2}−14x+49$ | $(x−7)_{2}$ |

$x_{2}−10x+25$ | $(x−5)_{2}$ |

$x_{2}+20x+100$ | $(x+10)_{2}$ |

When completing the square for a quadratic expression in the form $x_{2}+bx+c,$ the constant $c$ must also be taken into account. When adding and subtracting the term $(2b )_{2},$ the quadratic expression gets a constant term of $-(2b )_{2}+c.$ $x_{2}+bx+c⇓perfect square trinomialx_{2}+bx+(2b )_{2} −constant(2b )_{2}+c $

After completing its square, a quadratic expression can be rewritten into vertex form by factoring the perfect square trinomial. $perfect square trinomialx_{2}+bx+(2b )_{2} −(2b )_{2}+c=(x+2b )_{2}−(2b )_{2}+c$

Completing the square can be used when solving quadratic equations. An equation in the form $x_{2}+bx+c=0$ is then rewritten by isolating the $x_{2}-$ and $x-$terms. $x_{2}+bx+c=0⇓x_{2}+bx=-c $ By adding the term $(2b )_{2}$ to both sides of the equation, a perfect square trinomial is formed. $x_{2}+bx=-c⇓x_{2}+bx+(2b )_{2}=-c+(2b )_{2} $ The equation can then be rewritten by factoring the perfect square trinomial. $x_{2}+bx+(2b )_{2}=-c+(2b )_{2}⇓(x+2b )_{2}=-c+(2b )_{2} $ When the equation is written on this form it can be solved with square roots.

Complete the square

This is a method by which a quadratic expression is rewritten as a difference of a perfect square trinomial and a constant.
$x_{2}+bx+c⇓square(x+2b )_{2} −constant(2b )_{2}+c $
To use the Completing the Square method, there are five steps to follow.
### 1

It is easier to complete the square when the expression is written in the form $x_{2}+bx+c.$ Therefore, any coefficient of $x_{2}$ that does not equal $1$ should be factored out. $ax_{2}+bx+c⇔a(x_{2}+ab x+ac ) $ For simplicity of the following steps, it will be assumed that $a=1.$ For values of $a$ other than $1,$ the same steps should be performed. The only difference is that the new coefficients — $ab $ instead of $b$ and $ac $ instead of $c$ — will be used.

### 2

The constant needed to complete the square can now be identified by focusing on the $x_{2}-$ and $x-$terms, while ignoring the rest. One way to find it is by squaring half the coefficient of the $x-$term, which in this case is $b.$ $(2b )_{2} $ Note that leaving the constant as a power makes the next steps easier to perform.

### 3

The square can now be completed by adding and subtracting the constant found in Step $2.$ Note that the value of the original expression will not be changed, since the result of adding and subtracting the same value is equal to $0.$ $x_{2}+bx+c⇕perfect square trinomialx_{2}+bx+(2b )_{2} −constant(2b )_{2}+c $ The first three terms form a perfect square trinomial, which can be factored. The rest two terms do not contain the variable $x,$ so their value is constant.

### 4

The perfect square trinomial can now be factored and written as the square of a binomial.
The process of completing the square is now finished.
### 5

Finally, if needed, the expression can be simplified. In case $a$ was not equal to $1,$ now is a good time to remove the parentheses and multiply the obtained expression by $a.$
The method of completing the square is often used to solve quadratic equations. To do so, the perfect square polynomial should be written on one side of the equation, while the constant should be on another side. $x_{2}+bx+c=0⇕(x+2b )_{2}=(2b )_{2}−c $
Then the solutions are found by calculating the square roots of each side. ### Why

Factor the Coefficient of $x_{2}$

Identify the Constant Needed to Complete the Square

Complete the Square

Factor the Perfect Square Trinomial

$x_{2}+bx+(2b )_{2}−(2b )_{2}+c$

$x_{2}+2x(2b )+(2b )_{2}−(2b )_{2}+c$

FacPosPerfectSquare$a_{2}+2ab+b_{2}=(a+b)_{2}$

$(x+2b )_{2}−(2b )_{2}+c$

Simplify the expression

$a(x_{2}+ab x+ac )$

Complete the square

IdPropAddIdentity Property of Addition

$a(x_{2}+ab x+0+ac )$

$a(x_{2}+ab x+(2ab )_{2}−(2ab )_{2}+ac )$

$a(x_{2}+2x2ab +(2ab )_{2}−(2ab )_{2}+ac )$

FacPosPerfectSquare$a_{2}+2ab+b_{2}=(a+b)_{2}$

$a((x+2ab )_{2}−(2ab )_{2}+ac )$

Distribute $a$

PowQuot$(ba )_{m}=b_{m}a_{m} $

$a((x+2ab )_{2}−4a_{2}b_{2} +ac )$

DistrDistribute $a$

$a(x+2ab )_{2}−a⋅4a_{2}b_{2} +a⋅ac $

DenomMultFracToNumber$x⋅xa =a$

$a(x+2ab )_{2}−4ab_{2} +c$

Instead of a purely algebraic approach, it can be helpful to visualize the process of completing the square geometrically. Press the button below to see an animation of this method.

Complete the square

As it can be observed, the process of completing the square has the geometric meaning of finding a square that transforms a geometric figure compound of a square and a rectangle into another square.

Consider the quadratic function $f(x)=x_{2}−x−0.75.$ Complete the square to determine the vertex and zeros of the parabola. Then, use them to draw the graph.

Show Solution

To begin, we'll complete the square on $f$ by focusing on the coefficient of $x,$ which is $-1.$ Since $(2-1 )_{2}=0.25,$ $0.25$ is the constant term that will create a perfect square trinomial.
Notice that $f$ is now written in vertex form. It can be seen that $f$'s vertex is $(0.5,-1).$ To find the zeros we can set $f(x)$ equal to $0$ and solve.
Thus, the zeros of the parabola are $x=-0.5$ and $x=1.5.$ Lastly, we can use the vertex and the zeros to draw the graph of $f.$

$f(x)=x_{2}−x−0.75$

CompleteSquareComplete the square

$f(x)=(x_{2}−x+0.25)−0.75−0.25$

FacNegPerfectSquare$a_{2}−2ab+b_{2}=(a−b)_{2}$

$f(x)=(x−0.5)_{2}−0.75−0.25$

SubTermSubtract term

$f(x)=(x−0.5)_{2}−1$

$0=(x−0.5)_{2}−1$

AddEqn$LHS+1=RHS+1$

$1=(x−0.5)_{2}$

RearrangeEqnRearrange equation

$(x−0.5)_{2}=1$

SqrtEqn$LHS =RHS $

$x−0.5=±1$

AddEqn$LHS+0.5=RHS+0.5$

$x=0.5±1$

StateSolState solutions

$x_{1}=-0.5x_{2}=1.5 $

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