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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.
There are two general types of perfect square trinomials that can be useful when dealing with quadratic functions or quadratic equations. This process involves factoring the quadratic trinomial and rewriting the perfect square trinomial as the square of a binomial.

For a trinomial to be factorable as a perfect square trinomial, the first and last terms must be perfect squares and the middle term must be two times the square roots of the first and last terms. Consider the following expression.
To factor this trinomial, there three steps.
### 1

One good way to recognize if a trinomial is perfect square is to look at its first and last term. If they are both perfect squares, there's a good chance that it may be a perfect square trinomial. In the given expression, first and last terms can be written as the squares of 4x and 11, respectively.
### 2

The next step is to check whether the middle term is two times the square roots of the first and last terms.
### 3

Examine the First and Last Terms

$16x_{2}+88x+121⇓(4x)_{2}+88x+(11)_{2} $

These perfect squares show that the expression can be a perfect square trinomial. However, this is not enough to decide yet.
Examine the Middle Term

$16x_{2}+88x+121⇓(4x)_{2}+2(4x)(11)+(11)_{2} $

It can be seen that the given expression satisfies this condition, as well.
Write as a Square of a Binomial

Since the expression satisfies both conditions, it is a perfect square trinomial. Therefore, it can be written as a square of a binomial where $4x$ and $11$ are the first and second terms of the binomial, respectively.

$16x_{2}+88x+121=(4x+11)_{2} $

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 |
---|---|

x2−2x+1 | (x__ ___)2 |

x2−___x+___ | (x−7)2 |

x2−10x+___ | (x__ ___)2 |

x2+___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

a2−2ab+b2=(a−b)2

To help us identify the terms in the rule in the first example, we can for the expression x2−2x+1 rewrite 1 as 12 and 2x as 2⋅x⋅1.

x2−2⋅x⋅1+12=(x__ ___)2

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

x2−2⋅x⋅1+12=(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.

x2−___x+___ =(x−7)2

To fill in the second blank we use that b must be 7, which gives us that b2=72. When we study the first blank we can match a with x. What is left is 2⋅b=2⋅7.

x2−2⋅x⋅7+72=(x−7)2

This we prefer writing as x2−14x+49=(x−7)2. Let's continue with the third row

x2−10x+___=(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.

x2−10x+52=(x−5)2

By squaring 5 we get that the second row reads x2−10x+25=(x−5)2. We are now going to deal with the last row in the table.

x2+___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.

x2+20x+100=(x+10)2

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

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

x2−2x+1 | (x−1)2 |

x2−14x+49 | (x−7)2 |

x2−10x+25 | (x−5)2 |

x2+20x+100 | (x+10)2 |

$perfect square trinomialx_{2}+bx+(2b )_{2} −constant(2b )_{2} $

The quadratic expression can then be factored and written and written in vertex form.
$perfect square trinomialx_{2}+bx+(2b )_{2} −(2b )_{2}=(x+2b )_{2}−(2b )_{2}$

$x_{2}+bx+c⇓perfect square trinomialx_{2}+bx+(2b )_{2} −constant(2b )_{2}+c $

$perfect square trinomialx_{2}+bx+(2b )_{2} −(2b )_{2}+c=(x+2b )_{2}−(2b )_{2}+c$

Complete the square

This is a method by which a quadratic expression is rewritten as the difference of the square of a binomial and a constant.
### 1

It is easier to complete the square when the expression is written in the form x2±bx+c. Therefore, if the coefficient of x2 is not 1, it should be factored out.
For the 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-term, while ignoring the rest. One way to find this constant is by squaring half the coefficient of the x-term, which in this case is b.
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.
### 4

The perfect square trinomial can now be factored and rewritten 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.
### Why

$x_{2}±bx+c⇓square(x±2b )_{2} −constant(2b )_{2}+c $

To complete the square, there are five steps to follow.
Factor Out the Coefficient of x2

Identify the Constant Needed to Complete the Square

Complete the Square

$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 as the square of a binomial. The other two terms do not contain the variable x, so their value is constant.
Factor the Perfect Square Trinomial

$x_{2}±bx+(2b )_{2}−(2b )_{2}+c$

DenomMultFracToNumber

$2⋅2a =a$

$x_{2}±2(2b )x+(2b )_{2}−(2b )_{2}+c$

CommutativePropMult

Commutative Property of Multiplication

$x_{2}±2x(2b )+(2b )_{2}−(2b )_{2}+c$

a2±2ab+b2=(a±b)2

$(x±2b )_{2}−(2b )_{2}+c$

Simplify the expression

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

Derive function

IdPropAdd

Identity Property of Addition

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

Rewrite

Rewrite 0 as $(2ab )_{2}−(2ab )_{2}$

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

Rewrite

Rewrite $ab x$ as $2x2ab $

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

FacPosPerfectSquare

a2+2ab+b2=(a+b)2

$a((x+2ab )_{2}−(2ab )_{2}+ac )$

Simplify

PowQuot

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

$a((x+2ab )_{2}−4a_{2}b_{2} +ac )$

Distr

Distribute a

$a(x+2ab )_{2}−a⋅4a_{2}b_{2} +a⋅ac $

CancelCommonFac

Cancel out common factors

SimpQuotProd

Simplify quotient and product

$a(x+2ab )_{2}−4ab_{2} +c$

This method is usually extended to solve quadratic equations.

Instead of a purely algebraic approach, it can be helpful to visualize the process of completing the square geometrically.
*completes* an incomplete square.

The animation shows that the process of completing the square can be understood as finding the dimensions of the $square$ that

Consider the quadratic function

f(x)=x2−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)=x2−x−0.75

CompleteSquare

Complete the square

$f(x)=(x_{2}−x+0.25)−0.75−0.25$

FacNegPerfectSquare

a2−2ab+b2=(a−b)2

f(x)=(x−0.5)2−0.75−0.25

SubTerm

Subtract term

f(x)=(x−0.5)2−1

0=(x−0.5)2−1

AddEqn

LHS+1=RHS+1

1=(x−0.5)2

RearrangeEqn

Rearrange equation

(x−0.5)2=1

SqrtEqn

$LHS =RHS $

x−0.5=±1

AddEqn

LHS+0.5=RHS+0.5

x=0.5±1

StateSol

State solutions

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

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