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8.3: Completing the Square


In Introduction to Radical Notation, we showed how to solve equations such as (x^2 = 9) both algebraically and graphically.

[egin{aligned} x^{2} &=9 x &=pm 3 end{aligned} onumber ]

Note that when we take the square root of both sides of this equation, there are two answers, one negative and one positive.

A perfect square is nice, but not required. Indeed, we may even have to factor out a perfect square to put our final answer in simple form.

[egin{aligned} x^{2} &=8 x &=pm sqrt{8} x &=pm sqrt{4} sqrt{2} x &=pm 2 sqrt{2} end{aligned} onumber ]

Readers should use their calculators to check that (-2 sqrt{2} approx -2.8284) and (2 sqrt{2} approx 2.8284).

Now, let’s extend this solution technique to a broader class of equations.

Example (PageIndex{1})

Solve for (x : (x-4)^{2}=9)

Solution

Much like the solutions of (x^2 = 9) are (x = ±3), we use a similar approach on ((x−4)^2 = 9) to obtain:

[egin{array}{rlrl}{(x-4)^{2}} & {=9} & {} & color {Red} { ext { Original equation. }} {x-4} & {=pm 3} & {} & color {Red} { ext { There are two square roots. }}end{array} onumber ]

To complete the solution, add (4) to both sides of the equation.

[x=4 pm 3 quad color {Red} ext { Add } 3 ext { to both sides. } onumber ]

Note that this means that there are two answers, namely:

[egin{array}{l}{x=4-3} {x=1}end{array} onumber ]

or

[egin{array}{l}{x=4+3} {x=7}end{array} onumber ]

Check: Check each solution by substituting it into the original equation.

Substitute (1) for (x):

[egin{aligned}(x-4)^{2} &=9 (1-4)^{2} &=9 (-3)^{2} &=9 end{aligned} onumber ]

Substitute (7) for (x):

[egin{aligned}(x-4)^{2} &=9 (7-4)^{2} &=9 (3)^{2} &=9 end{aligned} onumber ]

Because the last statement in each check is a true statement, both (x = 1) and (x = 7) are valid solutions of ((x−4)^2 = 9).

Exercise (PageIndex{1})

Solve for (x :(x+6)^{2}=10)

Answer

(-2), (-10)

In Example (PageIndex{1}), the right-hand side of the equation ((x−4)^2 = 9) was a perfect square. However, this is not required, as the next example will show.

Example (PageIndex{2})

Solve for (x :(x+5)^{2}=7)

Solution

Using the same technique as in Example (PageIndex{1}), we obtain:

[egin{array}{rlrl}{(x+5)^{2}} & {=7} & {} & color {Red} { ext { Original equation. }} {x+5} & {=pm sqrt{7}} & {} & color {Red} { ext { There are two square roots. }}end{array} onumber ]

To complete the solution, subtract 5 from both sides of the equation.

[x=-5 pm sqrt{7} quad color {Red} ext { Subtract } 5 ext { from both sides.} onumber ]

Note that this means that there are two answers, namely:

[x=-5-sqrt{7} quad ext { or } quad x=-5+sqrt{7} onumber ]

Check: Check each solution by substituting it into the original equation.

Substitute (-5-sqrt{7}) for (x):

[egin{aligned}(x+5)^{2} &=7 ((-5-sqrt{7})+5)^{2} &=7 (-sqrt{7})^{2} &=7 end{aligned} onumber ]

Substitute (-5+sqrt{7}) for (x):

[egin{aligned}(x+5)^{2} &=7 ((-5+sqrt{7})+5)^{2} &=7 (sqrt{7})^{2} &=7 end{aligned} onumber ]

Because the last statement in each check is a true statement, both (x=-5-sqrt{7}) and (x=-5+sqrt{7}) are valid solutions of ((x+5)^{2}=7).

Exercise (PageIndex{2})

Solve for (x :(x-4)^{2}=5)

Answer

(4+sqrt{5}, 4-sqrt{5})

Sometimes you will have to factor out a perfect square to put your answer in simple form.

Example (PageIndex{3})

Solve for (x :(x+4)^{2}=20)

Solution

Using the same technique as in Example (PageIndex{1}), we obtain:

[egin{array}{rlrl}{(x+4)^{2}} & {=20} & {} & color {Red} { ext { Original equation. }} {x+4} & {=pm sqrt{20}} & {} & color {Red} { ext { There are two square roots. }} {x+4} & {=pm sqrt{4} sqrt{5}} & {} & color {Red} { ext { Factor out a perfect square. }} {x+4} & {=pm 2 sqrt{5}} & {} & color {Red} { ext { Simplify: } sqrt{4}=2}end{array} onumber ]

To complete the solution, subtract (4) from both sides of the equation.

[x=-4 pm 2 sqrt{5} quad color {Red} ext { Subtract } 4 ext { from both sides. } onumber ]

Note that this means that there are two answers, namely:

[x=-4-2 sqrt{5} quad ext { or } quad x=-4+2 sqrt{5} onumber ]

Check: Although it is possible to check the exact answers, let’s use our calculator instead. First, store (-4-2 sqrt{5}) in (mathbf{X}). Next, enter the left-hand side of the equation ((x + 4)^2 = 20) (see image on the left in Figure (PageIndex{3})). Note that (x+4)2 simplifies to 20, showing that (-4-2 sqrt{5}) is a solution of ((x+4)^2 = 20).

In similar fashion, the solution (-4+2 sqrt{5}) also checks in ((x + 4)^2 = 20) (see image on the right in Figure (PageIndex{3})).

Exercise (PageIndex{3})

Solve for (x :(x+7)^{2}=18)

Answer

(-7+3 sqrt{2},-7-3 sqrt{2})

Perfect Square Trinomials Revisited

Recall the squaring a binomial shortcut.

Squaring a Binomial

If (a) and (b) are any real numbers, then: [(a±b)^2 = a^2 ±2ab + b^2 onumber ]That is, you square the first term, take the product of the first and second terms and double the result, then square the third term.

Reminder examples:

[egin{aligned}(x+3)^{2} &=x^{2}+2(x)(3)+3^{2} &=x^{2}+6 x+9 end{aligned} onumber ]

[egin{aligned}(x-8)^{2} &=x^{2}-2(x)(8)+8^{2} &=x^{2}-16 x+64 end{aligned} onumber ]

Because factoring is “unmultiplying,” it is a simple matter to reverse the multiplication process and factor these perfect square trinomials.

[x^{2}+6 x+9=(x+3)^{2} onumber ]

[x^{2}-16 x+64=(x-8)^{2} onumber ]

Note how in each case we simply take the square root of the first and last terms.

Example (PageIndex{4})

Factor each of the following trinomials:

  1. (x^{2}-12 x+36)
  2. (x^{2}+10 x+25)
  3. (x^{2}-34 x+289)

Solution

Whenever the first and last terms of a trinomial are perfect squares, we should suspect that we have a perfect square trinomial.

  1. The first and third terms of (x^{2}-12 x+36) are perfect squares. Hence, we take their square roots and try:[x^{2}-12 x+36=(x-6)^{2} onumber ]Note that (2(x)(6)=12 x), which is the middle term on the left. The solution checks.
  2. The first and third terms of (x^{2}+10 x+25) are perfect squares. Hence, we take their square roots and try:[x^{2}+10 x+25=(x+5)^{2} onumber ]Note that (2(x)(5)=10 x), which is the middle term on the left. The solution checks.
  3. The first and third terms of (x^{2}-34 x+289) are perfect squares. Hence, we take their square roots and try:[x^{2}-34 x+289=(x-17)^{2} onumber ]Note that (2(x)(17)=34 x), which is the middle term on the left. The solution checks.

Exercise (PageIndex{4})

Factor: (x^2 + 30x + 225)

Answer

((x+15)^{2})

Completing the Square

In this section we start with the binomial (x^2 +bx) and ask the question “What constant value should we add to (x^2 + bx) so that the resulting trinomial is a perfect square trinomial?” The answer lies in this procedure.

Completing the square

To calculate the constant required to make (x^2 +bx) a perfect square trinomial:

  1. Take one-half of the coefficient of (x : dfrac{b}{2})
  2. Square the result of step one: (left(dfrac{b}{2} ight)^{2}=dfrac{b^{2}}{4})
  3. Add the result of step two to (x^{2}+b x : x^{2}+b x+dfrac{b^{2}}{4})

If you follow this process, the result will be a perfect square trinomial which will factor as follows:

[x^{2}+b x+dfrac{b^{2}}{4}=left(x+dfrac{b}{2} ight)^{2} onumber ]

Example (PageIndex{5})

Given (x^2 + 12 x), complete the square to create a perfect square trinomial.

Solution

Compare (x^2 + 12x) with (x^2 + bx) and note that (b = 12).

  1. Take one-half of (12: 6)
  2. Square the result of step one: (6^2 = 36)
  3. Add the result of step two to (x^2 + 12x: x^2 + 12x + 36)

Check: Note that the first and last terms of (x^2 +12x+36) are perfect squares. Take the square roots of the first and last terms and factor as follows:

[x^{2}+12 x+36=(x+6)^{2} onumber ]

Note that (2(x)(6) = 12x), so the middle term checks.

Exercise (PageIndex{5})

Given (x^2 + 16x), complete the square to create a perfect square trinomial.

Answer

(x^{2}+16 x+64=(x+8)^{2})

Example (PageIndex{6})

Given (x^2−3x), complete the square to create a perfect square trinomial.

Solution

Compare (x^2 −3x) with (x^2 + bx) and note that (b =−3).

  1. Take one-half of (-3 : -dfrac{3}{2})
  2. Square the result of step one: (left(-dfrac{3}{2} ight)^{2}=dfrac{9}{4})
  3. Add the result of step two to (x^{2}-3 x : x^{2}-3 x+dfrac{9}{4})

Check: Note that the first and last terms of (x^{2}-3 x+dfrac {9}{4}) are perfect squares. Take the square roots of the first and last terms and factor as follows:

[x^{2}-3 x+dfrac{9}{4}=left(x-dfrac{3}{2} ight)^{2} onumber ]

Note that (2(x)left (dfrac {3}{2} ight)=3 x), so the middle term checks.

Exercise (PageIndex{6})

Given (x^2 −5x), complete the square to create a perfect square trinomial.

Answer

(x^{2}-5 x+dfrac {10}{4}=left(x-dfrac {5}{2} ight)^{2})

Solving Equations by Completing the Square

Consider the following nonlinear equation.

[x^2 =2x +2 onumber ]

The standard approach is to make one side zero and factor.[x^2 −2x−2=0 onumber ] However, one quickly realizes that there is no integer pair whose product is (ac = −2) and whose sum is (b = −2). So, what does one do in this situation? The answer is “Complete the square.”

Example (PageIndex{7})

Use completing the square to help solve (x^2 =2x + 2).

Solution

First, move (2x) to the left-hand side of the equation, keeping the constant (2) on the right-hand side of the equation.[x^2 −2x =2 onumber ]On the left, take one-half of the coefficient of (x: left (dfrac{1}{2} ight)(-2)=-1). Square the result: ((-1)^{2}=1). Add this result to both sides of the equation.

[egin{array}{l}{x^{2}-2 x+1=2+1} {x^{2}-2 x+1=3}end{array} onumber ]

We can now factor the left-hand side as a perfect square trinomial.

[(x-1)^{2}=3 onumber ]

Now, as in Examples (PageIndex{1}), (PageIndex{2}), and (PageIndex{3}), we can take the square root of both sides of the equation. Remember, there are two square roots.

[x-1=pm sqrt{3} onumber ]

Finally, add (1) to both sides of the equation.

[x=1 pm sqrt{3} onumber ]

Thus, the equation (x^2 =2x+ 2) has two answers, (x=1-sqrt{3}) and (x=1+sqrt{3}).

Check: Let’s use the calculator to check the solutions. First, store (1-sqrt{3}) in (mathbf{X}) (see the image on the left in Figure (PageIndex{4})). Then enter the left- and right-hand sides of the equation (x^2 =2 x + 2) and compare the results (see the image on the left in Figure (PageIndex{4})). In similar fashion, check the second answer (1+sqrt{3}) (see the image on the right in Figure (PageIndex{4})).

In both cases, note that the left- and right-hand sides of (x^2 =2x+2) produce the same result. Hence, both (1-sqrt{3}) and (1+sqrt{3}) are valid solutions of (x^2 =2x+2).

Exercise (PageIndex{7})

Use completing the square to help solve (x^2 =3−6x).

Answer

(-3+2 sqrt{3},-3-2 sqrt{3})

Example (PageIndex{8})

Solve the equation (x^2 −8x−12 = 0), both algebraically and graphically. Compare your answer from each method.

Solution

First, move the constant (12) to the right-hand side of the equation.

[egin{aligned} x^{2}-8 x-12=0 & quad color {Red} ext { Original equation. } x^{2}-8 x=12 & quad color {Red} ext { Add } 12 ext { to both sides. } end{aligned} onumber ]

Take half of the coefficient of (x :(1 / 2)(-8)=-4). Square: ((-4)^{2}=16). Now add (16) to both sides of the equation.

[egin{aligned} x^{2}-8 x+16 & =12+16 quad color {Red} ext { Add } 16 ext { to both sides. } (x-4)^{2} & =28 quad color {Red} ext { Factor left-hand side. } x-4 &=pm sqrt{28} quad color {Red} ext { There are two square roots. }end{aligned} onumber ]
Note that the answer is not in simple radical form.

[egin{array}{rlrl}{x-4} & {=pm sqrt{4} sqrt{7}} & {} & color {Red} { ext { Factor out a perfect square. }} {x-4} & {=pm 2 sqrt{7}} & {} & color {Red} { ext { Simplify: } sqrt{4}=2} {x} & {=4 pm 2 sqrt{7}} & {} & color {Red} { ext { Add } 4 ext { to both sides. }}end{array} onumber ]

Graphical solution: Enter the equation (y = x^2 − 8x − 12) in (mathbf{Y1}) of the Y= menu (see the first image in Figure (PageIndex{5})). After some experimentation, we settled on the WINDOW parameters shown in the middle image of Figure (PageIndex{5}). Once you’ve entered these WINDOW parameters, push the GRAPH button to produce the rightmost image in Figure (PageIndex{5}).

We’re looking for solutions of (x^2 −8x−12 = 0), so we need to locate where the graph of (y = x^2 −8x−12) intercepts the (x)-axis. That is, we need to find the zeros of (y = x^2 −8x−12). Select 2:zero from the CALC menu, move the cursor slightly to the left of the first (x)-intercept and press ENTER in response to “Left bound.” Move the cursor slightly to the right of the first (x)-intercept and press ENTER in response to “Right bound.” Leave the cursor where it sits and press ENTER in response to “Guess.” The calculator responds by finding the (x)-coordinate of the (x)-intercept, as shown in the first image in Figure (PageIndex{6}).

Repeat the process to find the second (x)-intercept of (y = x^2−8x−12) shown in the second image in Figure (PageIndex{6}).

Reporting the solution on your homework: Duplicate the image in your calculator’s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.

  • Label the horizontal and vertical axes with (x) and (y), respectively (see Figure (PageIndex{7})).
  • Place your WINDOW parameters at the end of each axis (see Figure (PageIndex{7})).
  • Label the graph with its equation (see Figure (PageIndex{7})).
  • Drop dashed vertical lines through each (x)-intercept. Shade and label the (x)-values of the points where the dashed vertical line crosses the (x)-axis. These are the solutions of the equation (x^2−8x−12 = 0) (see Figure (PageIndex{7})).

Thus, the graphing calculator reports that the solutions of (x^2 −8x−12 = 0) are (x approx-1.291503) and (x approx 9.2915026).

Comparing exact and calculator approximations: How well do the graphing calculator solutions compare with the exact solutions, (x=4-2 sqrt{7}) and (x=4+2 sqrt{7})? After entering each in the calculator (see Figure (PageIndex{8})), the comparison is excellent!

Exercise (PageIndex{8})

Solve the equation (x^2 +6x + 3 = 0) both algebraically and graphically, then compare your answers.

Answer

(-3-sqrt{6},-3+sqrt{6})


Watch the video: Integrated Math: Completing the Square (November 2021).