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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.*

A perfect square trinomial is a trinomial that 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.

Perfect Square Trinomial | Square of Binomial |
---|---|

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

a2−2ab+b2 | (a−b)2 |

Once a quadratic expression is recognized as a perfect square trinomial, it can be written as a square of a binomial by factoring the perfect square trinomial.

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 are three steps.
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1

Confirm That the First and Last Terms Are Perfect Squares

One good way to recognize if a trinomial is a perfect square trinomial is to look at its first and last terms. If they are both perfect squares, there is a good chance that it is a perfect square trinomial. In the given expression, the first and last terms can be written as the squares of 4x and 11, respectively.
These perfect squares show that the expression could be a perfect square trinomial. However, this is not enough to decide yet.

2

Confirm That the Middle Term Is Twice the Product of the Square Roots of First and Last Terms

The next step is to check whether the middle term is two times the square roots of the first and last terms.
It can be seen that the given expression satisfies this condition as well.

3

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.

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 *expand_more*

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

In a perfect square trinomial, there is a relationship between the coefficient of the x-term and the constant term — the constant term is equal to the square of half the coefficient of the x-term.
This relationship can be used to form a perfect square trinomial by adding a constant c to *any* expression in the form x2+bx.
The process of finding the constant c can be visualized by using *algebraic tiles*. Consider the following expression.
The expression is represented using algebraic tiles. Then, a square is created by rearranging the existing tiles and adding more tiles. The following applet summarizes this process.
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This process is called completing the square. To complete the square for an expression algebraically, these steps can be followed.

1

Identify the Coefficient of the x-Term

For the given expression, the value of b is 6.

2

Calculate $(2b )_{2}$

Once the value of b is identified, calculate the square of half of the value of b.

3

Add $(2b )_{2}$ to the Initial Expression

Add $(2b )_{2}$ to the expression to obtain a perfect square trinomial.
In this case, 9 should be added to x2+6x.

4

Factor the Perfect Square Trinomial

The expression obtained in the previous step can be now factored as the square of a binomial.
This will be applied to the expression x2+6x+9.
Therefore, a perfect square trinomial is obtained by adding a constant $c=(2b )_{2}$ to the initial expression in the form x2+bx.

x2+6x+9

SplitIntoFactors

Split into factors

x2+2⋅3x+9

CommutativePropMult

Commutative Property of Multiplication

x2+2x⋅3+9

FacPosPerfectSquare

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

(x+3)2

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 *expand_more*

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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