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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 |
---|---|

$x_{2}−2x+1$ | $(x−1)_{2}$ |

$x_{2}−14x+49$ | $(x−7)_{2}$ |

$x_{2}−10x+25$ | $(x−5)_{2}$ |

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

Completing the square is a method by which a quadratic expression is rewritten as a difference of a perfect square trinomial and a constant. Commonly this is done by adding and subtracting a constant, $p.$ $x_{2}+bx=x_{2}+bx+=0p−p $ By choosing the constant, $p,$ as $(2b )_{2}$ the quadratic expression becomes a difference of a perfect square trinomial and a constant. $perfect square trinomialx_{2}+bx+(2b )_{2} −constant(2b )_{2} $ The quadratic expression can then be factored and written and written in vertex form. $perfect square trinomialx_{2}+bx+(2b )_{2} −(2b )_{2}=(x+2b )_{2}−(2b )_{2}$

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

All quadratic expressions in standard form can be rewritten by completing the square. One example of how this can be used in practice is to rewrite the function $f(x)=2x_{2}+12x+2 $ in vertex form to determine the vertex.

Factor the coefficient of $x_{2}$

Identify the constant needed to complete the square

Complete the square

Factor the perfect square trinomial

The perfect square trinomial can now be factored as normal.

$f(x)=2((x_{2}+6x+3_{2})−3_{2}+1)$

SplitIntoFactorsSplit into factors

$f(x)=2((x_{2}+2x⋅3+3_{2})−3_{2}+1)$

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

$f(x)=2((x+3)_{2}−3_{2}+1)$

Simplify the expression

Finally, the expression can be simplified. For the example, the goal is to write the function in vertex form to determine the vertex. To achieve this, the terms have to be simplified.

$f(x)=2((x+3)_{2}−3_{2}+1)$

CalcPowCalculate power

$f(x)=2((x+3)_{2}−9+1)$

AddTermsAdd terms

$f(x)=2((x+3)_{2}−8)$

DistrDistribute $2$

$f(x)=2(x+3)_{2}−16$

The function is now written in vertex form. It can be seen that the vertex is $(-3,-16).$

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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