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y 2 + 3 y .   The constant must be the square of one half the coefficient of y . Since the coefficient of y is 3, we have

3 2 and ( 3 2 ) 2 = 9 4
The constant is 9 4 .
y 2 + 3 y + 9 4 = ( y + 3 2 ) 2

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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 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.

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

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x 2 3 x 1 = 0. Add 1 to both sides . x 2 3 x = 1 One half the coefficient of  x  is  3 2 .  Square it:  ( 3 2 ) 2 = 9 4 . Add  9 4  to each side . x 2 3 x + 9 4 = 1 + 9 4 x 2 3 x + 9 4 = 13 4 Factor . Notice that since the sign of the middle  term of the trinomial is " " , its factored form has a " "  sign . ( x 3 2 ) 2 = 13 4 Now take square roots . x 3 2 = ± 13 4 x 3 2 = ± 13 2 x = ± 13 2 + 3 2 x = ± 13 + 3 2 x = 3 ± 13 2

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3 a 2 36 a 39 = 0. Add 39 to both sides . 3 a 2 36 a = 39 The leading coefficient is 3 and we  need it to be 1 . Divide each term by 3 . a 2 12 a = 13 One half the coefficient of  a  is  6. Square it:  ( 6 ) 2 =36 . Add 36 to each side . a 2 12 a + 36 = 13 + 36 a 2 12 a + 36 = 49 ( a 6 ) 2 = 49 Factor . a 6 = ± 7 a = ± 7 + 6 + 7 + 6 = 13 , 7 + 6 = 1 a = 13 , 1

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2 x 2 + x + 4 = 0 2 x 2 + x = 4 x 2 + 1 2 x = 2 x 2 + 1 2 x + ( 1 4 ) 2 = 2 + ( 1 4 ) 2 ( x + 1 4 ) 2 = 2 + 1 16 = 32 16 + 1 16 = 31 16
Since we know that the square of any number is positive, this equation has no real number solution.

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Use a calculator.  Calculator problem.  Solve 7 a 2 5 a 1 = 0. Round each solution to the nearest tenth.
7 a 2 5 a 1 = 0 7 a 2 5 a = 1 a 2 5 7 a = 1 7 a 2 5 7 a + ( 5 14 ) 2 = 1 7 + ( 5 14 ) 2 ( a 5 14 ) 2 = 1 7 + 25 196 = 28 196 + 25 196 = 53 196 a 5 14 = ± 53 196 = ± 53 14 a = 5 14 ± 53 14 = 5 ± 53 14

  1. We will first compute the value of the square root.

    Type 53 Press x Display reads: 7.2801099
    Press the key that places this value into memory.
  2. For a = 5 + 53 14 ,

    Type 5 Press + Press the key that recalls the value in memory . Press = Press ÷ Type 14 Press = Display reads: .87715071
    Rounding to tenths, we get a 0.9.
  3. For a = 5 53 14

    Type 5 Press Press the key that recalls the value in memory . Press = Press ÷ Type 14 Display reads: .16286499
    Rounding to tenths, we get a 0.2. Thus, a 0.9 and 0.2 to the nearest tenth.

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Practice set a

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

x 2 2 x 48 = 0

x = 6 , 8

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x 2 + 3 x 5 = 0

x = 3 ± 29 2

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4 m 2 + 5 m = 1

m = 1 4 , 1

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5 y 2 2 y 4 = 0

y = 1 ± 21 5

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Use a calculator.  Calculator problem.  Solve 3 x 2 x 1 = 0. Round each solution to the nearest tenth.

x = 0.8 , 0.4

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Exercises

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

x 2 + 2 x 8 = 0

x = 4 , 2

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a 2 + 7 a + 12 = 0

a = 3 , 4

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y 2 2 y 24 = 0

y = 4 , 6

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x 2 + 2 x + 5 = 0

No real number solution.

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x 2 + 4 x + 4 = 0

x = 2

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b 2 + 5 b 3 = 0

b = 5 ± 37 2

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a 2 + 10 a 9 = 0

a = 5 ± 34

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x 2 10 x = 0

x = 10 , 0

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a 2 6 a = 0

a = 6 , 0

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x 2 14 x = 13

x = 13 , 1

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2 a 2 + 2 a 1 = 0

a = 1 ± 3 2

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9 x 2 + 12 x 5 = 0

x = 1 3 , 5 3

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16 y 2 8 y 3 = 0

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2 x 2 + 5 x 4 = 0

x = 5 ± 57 4

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x 2 + 2 x + 8 = 0

No real number solution.

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7 a 2 + 3 a 1 = 0

a = 3 ± 37 14

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Use a calculator.  calculator problems

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

3 y 2 + 5 y = 7

y = 0.91 , 2.57

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1.8 x 2 + 2.3 x 4.1 = 0

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0.04 a 2 0.03 a + 0.02 = 0

No real number solution.

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Exercises for review

( [link] ) Factor 12 a x 6 b x + 20 a y 10 b y by grouping.

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( [link] ) Graph the compound inequality 6 2 x + 2 < 4.

A horizontal line with arrows on both ends.

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.

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

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( [link] ) Find the product: x 2 4 x 12 x 2 2 x 8 x 2 3 x 4 x 2 3 x 18 .

x + 1 x + 3

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( [link] ) Use the method of extraction of roots to solve ( x 2 ) 2 = 25.

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Questions & Answers

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Quantity demanded refers to the specific amount of a good or service that consumers are willing and able to purchase at a give price and within a specific time period. Demand, on the other hand, is a broader concept that encompasses the entire relationship between price and quantity demanded
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Source:  OpenStax, Elementary algebra. OpenStax CNX. May 08, 2009 Download for free at http://cnx.org/content/col10614/1.3
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