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10.5: Solving Quadratic Equations Using the Method of Completing the Square

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The Logic Behind The Method

Suppose we wish to solve the quadratic equation x23x1=0. Since the equation is not of the form x2=K, we cannot use extraction of roots. Next, we try factoring, but after a few trials we see that x23x1 is not factorable. We need another method for solving quadratic equations.

The method we shall study is based on perfect square trinomials and extraction of roots. The method is called solving quadratic equations by completing the square. Consider the equation

x2+6x+5=0.

This quadratic equation could be solved by factoring, but we'll use the method of completing the square. We will explain the method in detail after we look at this example. First we'll rewrite the equation as

x2+6x=5

Then, we'll add 9 to each side. We get

x2+6x+9=5+9

The left side factors as a perfect square trinomial.

(x+3)2=4

We can solve this by the extraction of roots.

x+3=±4x+3=±2x=±23x=+23 and x=23x=1 and 5

Notice that when the roots are rational numbers, the equation is factorable.

The big question is, "How did we know to add 9 to each side of the equation?" We can convert any quadratic trinomial appearing in an equation into a perfect square trinomial if we know what number to add to both sides. We can determine that particular number by observing the following situation:

Consider the square of the binomial and the resulting perfect square trinomial

(x+p)2=x2+2px+p2

Notice that the constant term (the number we are looking for) can be obtained from the linear term 2px. If we take one half the coefficient of x, 202=p, and square it, we get the constant term p2. This is true for every perfect square trinomial with a leading coefficient 1.

Note

In a perfect square trinomial with leading coefficient 1, the constant term is the square of one-half the coefficient of the linear term.

Study these examples to see what constant term will make the given binomial into a perfect square trinomial.

Example 10.5.1

x2+6x. The constant must be the square of one half the coefficient of x. Since the coefficient of x is 6, we have

62=3 and 32=9

The constant is 9.

x2+6x+9=(x+3)2

This is a perfect square trinomial.

Example 10.5.2

a2+10a. The constant must be the square of one half the coefficient of a. Since the coefficient of a is 10, we have

102=5 and 52=25.

The constant is 25

a2+10a+25=(a+5)2

Example 10.5.3

y2+3y. The constant must be the square of one-half the coefficient of y. Since the coefficient of y is 3, we have

32 and (32)2=94

The constant is 94.

y2+3y+94=(y+32)2

The Method Of Completing The Square

Now, with these observations, we can describe the method of completing the square.

The Method of Completing the Square
  1. Write the equation so that the constant term appears on the right side of the equation.
  2. If the leading coefficient is different from 1, divide each term of the equation by that coefficient.
  3. Take one half of the coefficient of the linear term, square it, then add it to both sides of the equation.
  4. The trinomial on the left is now a perfect square trinomial and can be factored as ()2. The first term in the parentheses is the square root of the quadratic term. The last term in the parentheses is one-half the coefficient of the linear term.
  5. Solve this equation by extraction of roots.

Sample Set A

Solve the following equations.

Example 10.5.4

x2+8x9=0 Add 9 to both sides.x2+8x=9 One half the coefficient of x is 4, and 42 is 16. Add 16 to both sides. x2+8x+16=9+16x2+8x+16=25 Factor. (x+4)2=25 Take square roots. x+4=±5x=±54+54=1,54=9x=1,9

Example 10.5.5

x23x1=0 Add 1 to both sides. One half the coefficient of x is 32.x23x=1 Square it: (32)2=94. Add 94to each side.x23x+94=1+94x23x+94=134Factor. Notice that since the sign of the middle term of the trinomial is "-", its factored form has a "-" sign.(x32)2=134 Now take square roots x32=±134x32=±132x=±132+32x=±13+32x=3±132

Example 10.5.6

3a236a39=0 Add 39 to both sides.3a236a=39 The leading coefficient is 3 and we need it to be 1. Divide each term by 3. One half of the coefficient of a is 6a212a=13 Square it: (6)2=36 Add 36 to each side. a212a+36=13+36a212a+36=49(a6)2=49 Factor. a6=±7a=±7+6+7+6=13,7+6=1

a=13,1.

Example 10.5.7

2x2+x+4=02x2+x=4x2+12x=2x2+12x+(14)2=2+(14)2

(x+14)2=2+116=3216+116=3116

Since we know that the square of any number is positive, this equation has no real number solution.

Example 10.5.8

Calculator Problem:

Solve 7a25a1=0. Round each solution to the nearest tenth.

7a25a1=07a25a=1a257a=17a257a+(514)2=17+(514)2(a514)2=17+25196=28196+25196=53196a514=±53196=±5314

a=514±5314=5±5314

Rounding to tenths, we get a0.2. Thus, a0.9 and 0.2 to the nearest tenth.

Practice Set A

Solve each of the following quadratic equations using the method of completing the square.

Practice Problem 10.5.1

x22x48=0

Answer

x=6,8

Practice Problem 10.5.2

x2+3x5=0

Answer

x=3±292

Practice Problem 10.5.3

4m2+5m=1

Answer

m=14,1

Practice Problem 10.5.4

5y22y4=0

Answer

y=1±215

Practice Problem 10.5.5

Calculator problem:

Solve 3x2x1=0. Round each solution to the nearest tenth.

Answer

x=0.8,0.4

Exercises

For the following problems, solve the equations by completing the square.

Exercise 10.5.1

x2+2x8=0

Answer

x=4,2

Exercise 10.5.2

y25y6=0

Exercise 10.5.3

a2+7a+12=0

Answer

a=3,4

Exercise 10.5.4

x210x+16=0

Exercise 10.5.5

y22y24=0

Answer

y=4,6

Exercise 10.5.6

a2+2a35=0

Exercise 10.5.7

x2+2x+5=0

Answer

No real number solution.

Exercise 10.5.8

x26x+1=0

Exercise 10.5.9

x2+4x+4=0

Answer

x=2

Exercise 10.5.10

a2+4a+7=0

Exercise 10.5.11

b2+5b3=0

Answer

b=5±372

Exercise 10.5.12

b26b=72

Exercise 10.5.13

a2+10a9=0

Answer

a=5±34

Exercise 10.5.14

a22a3=0

Exercise 10.5.15

x210x=0

Answer

x=10,0

Exercise 10.5.16

y28y=0

Exercise 10.5.17

a26a=0

Answer

a=6,0

Exercise 10.5.18

b2+6b=0

Exercise 10.5.19

x214x=13

Answer

x=13,1

Exercise 10.5.20

x2+8x=84

Exercise 10.5.21

2a2+2a1=0

Answer

a=1±32

Exercise 10.5.22

4b28b=16

Exercise 10.5.23

9x2+12x5=0

Answer

x=13,53

Exercise 10.5.24

16y28y3=0

Exercise 10.5.25

2x2+5x4=0

Answer

x=5±574

Exercise 10.5.26

3a2+2a24=0

Exercise 10.5.27

x2+2x+8=0

Answer

No real number solution.

Exercise 10.5.28

y23y+10=0

Exercise 10.5.29

7a2+3a1=0

Answer

a=3±3714

Calculator Problems

For the following problems, round each solution to the nearest hundredth.

Exercise 10.5.30

5m22m6=0

Exercise 10.5.31

3y2+5y=7

Answer

y=0.91,2.57

Exercise 10.5.32

1.8x2+2.3x4.1=0

Exercise 10.5.33

0.04a20.03a+0.02=0

Answer

No real number solution.

Exercises For Review

Exercise 10.5.34

Factor 12ax6bx+20ay10by by grouping.

Exercise 10.5.35

Graph the compound inequality 62x+2<4

A horizontal line with arrows on both ends.

Answer

A number line with arrows on each end, and labeled from negative two to five in increments of one. There is a closed circle at four and an open circle at three. These circles are connected by a black line.

Exercise 10.5.36

Find the equation of the line that passes through the points (1,2) and (0,4)

Exercise 10.5.37

Find the product: x24x12x22x8x23x4x23x18

Answer

x+1x+3

Exercise 10.5.38

Use the method of extraction of roots to solve (x2)2=25


This page titled 10.5: Solving Quadratic Equations Using the Method of Completing the Square is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Denny Burzynski & Wade Ellis, Jr. (OpenStax CNX) .

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