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6.1 Extracting Square Roots and Completing the Square

Learning Objectives

  1. Solve certain quadratic equations by extracting square roots.
  2. Solve any quadratic equation by completing the square.

Extracting Square Roots

Recall that a quadratic equation is in standard formAny quadratic equation in the form ax2+bx+c=0, where a, b, and c are real numbers and a0. if it is equal to 0:

ax2+bx+c=0 where a, b, and c are real numbers and a0. A solution to such an equation is a root of the quadratic function defined by f(x)=ax2+bx+c. Quadratic equations can have two real solutions, one real solution, or no real solution—in which case there will be two complex solutions. If the quadratic expression factors, then we can solve the equation by factoring. For example, we can solve 4x29=0 by factoring as follows:

4x29=0(2x+3)(2x3)=02x+3=0or2x3=02x=32x=3x=32x=32

The two solutions are ±32. Here we use ± to write the two solutions in a more compact form. The goal in this section is to develop an alternative method that can be used to easily solve equations where b=0, giving the form

ax2+c=0

The equation 4x29=0 is in this form and can be solved by first isolating x2.

4x29=04x2=9x2=94

If we take the square root of both sides of this equation, we obtain the following:

x2=94|x|=32

Here we see that x=±32 are solutions to the resulting equation. In general, this describes the square root propertyFor any real number k, if x2=k, then x=±k.; for any real number k,

ifx2=k,thenx=±k

Applying the square root property as a means of solving a quadratic equation is called extracting the rootApplying the square root property as a means of solving a quadratic equation.. This method allows us to solve equations that do not factor.

Example 1

Solve: 9x28=0.

Solution:

Notice that the quadratic expression on the left does not factor. However, it is in the form ax2+c=0 and so we can solve it by extracting the roots. Begin by isolating x2.

9x28=09x2=8x2=89

Next, apply the square root property. Remember to include the ± and simplify.

x=±89=±223

For completeness, check that these two real solutions solve the original quadratic equation.

Checkx=223

Checkx=223

9x28=09(223)28=09(429)8=088=00=0

9x28=09(223)28=09(429)8=088=00=0

Answer: Two real solutions, ±223

Sometimes quadratic equations have no real solution. In this case, the solutions will be complex numbers.

Example 2

Solve: x2+25=0.

Solution:

Begin by isolating x2 and then apply the square root property.

x2+25=0x2=25x=±25

After applying the square root property, we are left with the square root of a negative number. Therefore, there is no real solution to this equation; the solutions are complex. We can write these solutions in terms of the imaginary unit i=1.

x=±25=±125=±i5=±5i

Checkx=5i

Checkx=5i

x2+25=0(5i)2+25=025i2+25=025(1)+25=025+25=00=0

x2+25=0(5i)2+25=025i2+25=025(1)+25=025+25=00=0

Answer: Two complex solutions, ±5i.

Try this! Solve: 2x23=0.

Answer: The solutions are ±62.

Consider solving the following equation:

(x+5)2=9

To solve this equation by factoring, first square x+5 and then put the equation in standard form, equal to zero, by subtracting 9 from both sides.

(x+5)2=9x2+10x+25=9x2+10x+16=0

Factor and then apply the zero-product property.

x2+10x+16=0(x+8)(x+2)=0x+8=0orx+2=0x=8x=2

The two solutions are −8 and −2. When an equation is in this form, we can obtain the solutions in fewer steps by extracting the roots.

Example 3

Solve by extracting roots: (x+5)2=9.

Solution:

The term with the square factor is isolated so we begin by applying the square root property.

(x+5)2=9Applythesquarerootproperty.x+5=±9Simplify.x+5=±3x=5±3

At this point, separate the “plus or minus” into two equations and solve each individually.

x=5+3orx=53x=2x=8

Answer: The solutions are −2 and −8.

In addition to fewer steps, this method allows us to solve equations that do not factor.

Example 4

Solve: 2(x2)25=0.

Solution:

Begin by isolating the term with the square factor.

2(x2)25=02(x2)2=5(x2)2=52

Next, extract the roots, solve for x, and then simplify.

x2=±52Rationalizethedenominator.x=2±5222x=2±102x=4±102

Answer: The solutions are 4102 and 4+102.

Try this! Solve: 2(3x1)2+9=0.

Answer: The solutions are 13±22i.

Completing the Square

In this section, we will devise a method for rewriting any quadratic equation of the form

ax2+bx+c=0 as an equation of the form

(xp)2=q

This process is called completing the squareThe process of rewriting a quadratic equation to be in the form (xp)2=q.. As we have seen, quadratic equations in this form can be easily solved by extracting roots. We begin by examining perfect square trinomials:

(x+3)2=x2+6x+9(62)2=(3)2=9

The last term, 9, is the square of one-half of the coefficient of x. In general, this is true for any perfect square trinomial of the form x2+bx+c.

(x+b2)2=x2+2b2x+(b2)2=x2+bx+(b2)2

In other words, any trinomial of the form x2+bx+c will be a perfect square trinomial if

c=(b2)2

Note: It is important to point out that the leading coefficient must be equal to 1 for this to be true.

Example 5

Complete the square: x26x+?=(x+?)2.

Solution:

In this example, the coefficient b of the middle term is −6. Find the value that completes the square as follows:

(b2)2=(62)2=(3)2=9

The value that completes the square is 9.

x26x+9=(x3)(x3)=(x3)2

Answer: x26x+9=(x3)2

Example 6

Complete the square: x2+x+?=(x+?)2.

Solution:

Here b = 1. Find the value that will complete the square as follows:

(b2)2=(12)2=14

The value 14 completes the square:

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

Answer: x2+x+14=(x+12)2

We can use this technique to solve quadratic equations. The idea is to take any quadratic equation in standard form and complete the square so that we can solve it by extracting roots. The following are general steps for solving a quadratic equation with leading coefficient 1 in standard form by completing the square.

Example 7

Solve by completing the square: x28x2=0.

Solution:

It is important to notice that the leading coefficient is 1.

Step 1: Add or subtract the constant term to obtain an equation of the form x2+bx=c. Here we add 2 to both sides of the equation.

x28x2=0x28x=2

Step 2: Use (b2)2 to determine the value that completes the square. In this case, b = −8:

(b2)2=(82)2=(4)2=16

Step 3: Add (b2)2 to both sides of the equation and complete the square.

x28x=2x28x+16=2+16(x4)(x4)=18(x4)2=18

Step 4: Solve by extracting roots.

(x4)2=18x4=±18x=4±92x=4±32

Answer: The solutions are 432 and 4+32. The check is left to the reader.

Example 8

Solve by completing the square: x2+2x48=0.

Solution:

Begin by adding 48 to both sides.

x2+2x48=0x2+2x=48

Next, find the value that completes the square using b = 2.

(b2)2=(22)2=(1)2=1

To complete the square, add 1 to both sides, complete the square, and then solve by extracting the roots.

x2+2x=48Completethesquare.x2+2x+1=48+1(x+1)(x+1)=49(x+1)2=49Extracttheroots.x+1=±49x+1=±7x=1±7

At this point, separate the “plus or minus” into two equations and solve each individually.

x=17orx=1+7x=8x=6

Answer: The solutions are −8 and 6.

Note: In the previous example the solutions are integers. If this is the case, then the original equation will factor.

x2+2x48=0(x6)(x+8)=0

If an equation factors, we can solve it by factoring. However, not all quadratic equations will factor. Furthermore, equations often have complex solutions.

Example 9

Solve by completing the square: x210x+26=0.

Solution:

Begin by subtracting 26 from both sides of the equation.

x210x+26=0x210x=26

Here b = −10, and we determine the value that completes the square as follows:

(b2)2=(102)2=(5)2=25

To complete the square, add 25 to both sides of the equation.

x210x=26x210x+25=26+25x210x+25=1

Factor and then solve by extracting roots.

x210x+25=1(x5)(x5)=1(x5)2=1x5=±1x5=±ix=5±i

Answer: The solutions are 5±i.

Try this! Solve by completing the square: x22x17=0.

Answer: The solutions are x=1±32.

The coefficient of x is not always divisible by 2.

Example 10

Solve by completing the square: x2+3x+4=0.

Solution:

Begin by subtracting 4 from both sides.

x2+3x+4=0x2+3x=4

Use b = 3 to find the value that completes the square:

(b2)2=(32)2=94

To complete the square, add 94 to both sides of the equation.

x2+3x=4x2+3x+94=4+94(x+32)(x+32)=164+94(x+32)2=74

Solve by extracting roots.

(x+32)2=74x+32=±174x+32=±i72x=32±72i

Answer: The solutions are 32±72i.

So far, all of the examples have had a leading coefficient of 1. The formula (b2)2 determines the value that completes the square only if the leading coefficient is 1. If this is not the case, then simply divide both sides by the leading coefficient before beginning the steps outlined for completing the square.

Example 11

Solve by completing the square: 2x2+5x1=0.

Solution:

Notice that the leading coefficient is 2. Therefore, divide both sides by 2 before beginning the steps required to solve by completing the square.

2x2+5x12=022x22+5x212=0x2+52x12=0

Add 12 to both sides of the equation.

x2+52x12=0x2+52x=12

Here b=52, and we can find the value that completes the square as follows:

(b2)2=(5/22)2=(5212)2=(54)2=2516

To complete the square, add 2516 to both sides of the equation.

x2+52x=12x2+52x+2516=12+2516(x+54)(x+54)=816+2516(x+54)2=3316

Next, solve by extracting roots.

(x+54)2=3316x+54=±3316x+54=±334x=54±334x=5±334

Answer: The solutions are 5±334.

Try this! Solve by completing the square: 3x22x+1=0.

Answer: The solutions are x=13±23i.

Key Takeaways

  • Solve equations of the form ax2+c=0 by extracting the roots.
  • Extracting roots involves isolating the square and then applying the square root property. Remember to include “±” when taking the square root of both sides.
  • After applying the square root property, solve each of the resulting equations. Be sure to simplify all radical expressions and rationalize the denominator if necessary.
  • Solve any quadratic equation by completing the square.
  • You can apply the square root property to solve an equation if you can first convert the equation to the form (xp)2=q.
  • To complete the square, first make sure the equation is in the form x2+bx=c. The leading coefficient must be 1. Then add the value (b2)2 to both sides and factor.
  • The process for completing the square always works, but it may lead to some tedious calculations with fractions. This is the case when the middle term, b, is not divisible by 2.

Topic Exercises

    Part A: Extracting Square Roots

      Solve by factoring and then solve by extracting roots. Check answers.

    1. x216=0

    2. x236=0

    3. 9y21=0

    4. 4y225=0

    5. (x2)21=0

    6. (x+1)24=0

    7. 4(y2)29=0

    8. 9(y+1)24=0

    9. (u5)225=0

    10. (u+2)24=0

      Solve by extracting the roots.

    1. x2=81

    2. x2=1

    3. y2=19
    4. y2=116
    5. x2=12

    6. x2=18

    7. 16x2=9

    8. 4x2=25

    9. 2t2=1

    10. 3t2=2

    11. x240=0

    12. x224=0

    13. x2+1=0

    14. x2+100=0

    15. 5x21=0

    16. 6x25=0

    17. 8x2+1=0

    18. 12x2+5=0

    19. y2+4=0

    20. y2+1=0

    21. x249=0
    22. x2925=0
    23. x28=0

    24. t218=0

    25. x2+8=0

    26. x2+125=0

    27. 5y22=0

    28. 3x21=0

    29. (x+7)24=0

    30. (x+9)236=0

    31. (x5)220=0

    32. (x+1)228=0

    33. (3t+2)2+6=0

    34. (3t5)2+10=0

    35. 4(3x+1)227=0

    36. 9(2x3)28=0

    37. 2(3x1)2+3=0

    38. 5(2x1)2+2=0

    39. 3(y23)232=0
    40. 2(3y13)252=0
    41. 3(t1)2+12=0

    42. 2(t+1)2+8=0

    43. Solve for x: px2q=0, p,q>0

    44. Solve for x: (xp)2q=0, p,q>0

    45. The diagonal of a square measures 3 centimeters. Find the length of each side.

    46. The length of a rectangle is twice its width. If the diagonal of the rectangle measures 10 meters, then find the dimensions of the rectangle.

    47. If a circle has an area of 50π square centimeters, then find its radius.

    48. If a square has an area of 27 square centimeters, then find the length of each side.

    49. The height in feet of an object dropped from an 18-foot stepladder is given by h(t)=16t2+18, where t represents the time in seconds after the object is dropped. How long does it take the object to hit the ground? (Hint: The height is 0 when the object hits the ground. Round to the nearest hundredth of a second.)

    50. The height in feet of an object dropped from a 50-foot platform is given by h(t)=16t2+50, where t represents the time in seconds after the object is dropped. How long does it take the object to hit the ground? (Round to the nearest hundredth of a second.)

    51. How high does a 22-foot ladder reach if its base is 6 feet from the building on which it leans? Round to the nearest tenth of a foot.

    52. The height of a triangle is 12 the length of its base. If the area of the triangle is 72 square meters, find the exact length of the triangle’s base.

    Part B: Completing the Square

      Complete the square.

    1. x22x+?=(x?)2

    2. x24x+?=(x?)2

    3. x2+10x+?=(x+?)2

    4. x2+12x+?=(x+?)2

    5. x2+7x+?=(x+?)2

    6. x2+5x+?=(x+?)2

    7. x2x+?=(x?)2

    8. x212x+?=(x?)2

    9. x2+23x+?=(x+?)2

    10. x2+45x+?=(x+?)2

      Solve by factoring and then solve by completing the square. Check answers.

    1. x2+2x8=0

    2. x28x+15=0

    3. y2+2y24=0

    4. y212y+11=0

    5. t2+3t28=0

    6. t27t+10=0

    7. 2x2+3x2=0

    8. 3x2x2=0

    9. 2y2y1=0

    10. 2y2+7y4=0

      Solve by completing the square.

    1. x2+6x1=0

    2. x2+8x+10=0

    3. x22x7=0

    4. x26x3=0

    5. y22y+4=0

    6. y24y+9=0

    7. t2+10t75=0

    8. t2+12t108=0

    9. u223u13=0
    10. u245u15=0
    11. x2+x1=0

    12. x2+x3=0

    13. y2+3y2=0

    14. y2+5y3=0

    15. x2+3x+5=0

    16. x2+x+1=0

    17. x27x+112=0
    18. x29x+32=0
    19. t212t1=0
    20. t213t2=0
    21. 4x28x1=0

    22. 2x24x3=0

    23. 3x2+6x+1=0

    24. 5x2+10x+2=0

    25. 3x2+2x3=0

    26. 5x2+2x5=0

    27. 4x212x15=0

    28. 2x2+4x43=0

    29. 2x24x+10=0

    30. 6x224x+42=0

    31. 2x2x2=0

    32. 2x2+3x1=0

    33. 3u2+2u2=0

    34. 3u2u1=0

    35. x24x1=15

    36. x212x+8=10

    37. x(x+1)11(x2)=0

    38. (x+1)(x+7)4(3x+2)=0

    39. y2=(2y+3)(y1)2(y1)

    40. (2y+5)(y5)y(y8)=24

    41. (t+2)2=3(3t+1)

    42. (3t+2)(t4)(t8)=110t

      Solve by completing the square and round the solutions to the nearest hundredth.

    1. (2x1)2=2x

    2. (3x2)2=515x

    3. (2x+1)(3x+1)=9x+4

    4. (3x+1)(4x1)=17x4

    5. 9x(x1)2(2x1)=4x

    6. (6x+1)26(6x+1)=0

    Part C: Discussion Board

    1. Create an equation of your own that can be solved by extracting the roots. Share it, along with the solution, on the discussion board.

    2. Explain why the technique of extracting roots greatly expands our ability to solve quadratic equations.

    3. Explain why the technique for completing the square described in this section requires that the leading coefficient be equal to 1.

    4. Derive a formula for the diagonal of a square in terms of its sides.

Answers

  1. −4, 4

  2. 13,13

  3. 1, 3

  4. 12,72

  5. 0, 10

  6. ±9

  7. ±13
  8. ±23

  9. ±34
  10. ±22
  11. ±210

  12. ±i

  13. ±55
  14. ±24i
  15. ±2i

  16. ±23

  17. ±22

  18. ±2i2

  19. ±105
  20. −9, −5

  21. 5±25

  22. 23±63i
  23. 2±336
  24. 13±66i
  25. 4±326
  26. −1, 3

  27. x=±pqp

  28. 322 centimeters

  29. 52 centimeters

  30. 1.06 seconds

  31. 21.2 feet

  1. x22x+1=(x1)2

  2. x2+10x+25=(x+5)2

  3. x2+7x+494=(x+72)2
  4. x2x+14=(x12)2
  5. x2+23x+19=(x+13)2
  6. −4, 2

  7. −6, 4

  8. −7, 4

  9. 2,12

  10. 12,1

  11. 3±10

  12. 1±22

  13. 1±i3

  14. −15, 5

  15. 13,1

  16. 1±52
  17. 3±172
  18. 32±112i
  19. 7±332
  20. 1±174
  21. 2±52
  22. 3±63
  23. 1±103
  24. 3±262
  25. 1±2i

  26. 1±174
  27. 1±73
  28. 2±25

  29. 5±3

  30. 1±52
  31. 5±212
  32. 0.19, 1.31

  33. −0.45, 1.12

  34. 0.33, 0.67

  1. Answer may vary

  2. Answer may vary