$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 8.3: Completing the Square

• • Contributed by David Arnold
• Retired Professor (Mathematics) at College of the Redwoods
$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

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

\begin{aligned} x^{2} &=9 \\ x &=\pm 3 \end{aligned} \nonumber Figure $$\PageIndex{1}$$

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.

\begin{aligned} x^{2} &=8 \\ x &=\pm \sqrt{8} \\ x &=\pm \sqrt{4} \sqrt{2} \\ x &=\pm 2 \sqrt{2} \end{aligned} \nonumber Figure $$\PageIndex{2}$$

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:

$\begin{array}{rlrl}{(x-4)^{2}} & {=9} & {} & \color {Red} {\text { Original equation. }} \\ {x-4} & {=\pm 3} & {} & \color {Red} {\text { There are two square roots. }}\end{array} \nonumber$

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

$x=4 \pm 3 \quad \color {Red} \text { Add } 3 \text { to both sides. } \nonumber$

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

$\begin{array}{l}{x=4-3} \\ {x=1}\end{array} \nonumber$

or

$\begin{array}{l}{x=4+3} \\ {x=7}\end{array} \nonumber$

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

Substitute $$1$$ for $$x$$:

\begin{aligned}(x-4)^{2} &=9 \1-4)^{2} &=9 \\(-3)^{2} &=9 \end{aligned} \nonumber Substitute \(7 for $$x$$:

\begin{aligned}(x-4)^{2} &=9 \7-4)^{2} &=9 \\(3)^{2} &=9 \end{aligned} \nonumber 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 $$x :(x+6)^{2}=10$$

$$-2$$, $$-10$$

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:

$\begin{array}{rlrl}{(x+5)^{2}} & {=7} & {} & \color {Red} {\text { Original equation. }} \\ {x+5} & {=\pm \sqrt{7}} & {} & \color {Red} {\text { There are two square roots. }}\end{array} \nonumber$

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

$x=-5 \pm \sqrt{7} \quad \color {Red} \text { Subtract } 5 \text { from both sides.} \nonumber$

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

$x=-5-\sqrt{7} \quad \text { or } \quad x=-5+\sqrt{7} \nonumber$

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

Substitute $$-5-\sqrt{7}$$ for $$x$$:

\begin{aligned}(x+5)^{2} &=7 \(-5-\sqrt{7})+5)^{2} &=7 \\(-\sqrt{7})^{2} &=7 \end{aligned} \nonumber Substitute \(-5+\sqrt{7} for $$x$$:

\begin{aligned}(x+5)^{2} &=7 \(-5+\sqrt{7})+5)^{2} &=7 \\(\sqrt{7})^{2} &=7 \end{aligned} \nonumber 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 $$x :(x-4)^{2}=5$$

$$4+\sqrt{5}, 4-\sqrt{5}$$

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:

$\begin{array}{rlrl}{(x+4)^{2}} & {=20} & {} & \color {Red} {\text { Original equation. }} \\ {x+4} & {=\pm \sqrt{20}} & {} & \color {Red} {\text { There are two square roots. }} \\ {x+4} & {=\pm \sqrt{4} \sqrt{5}} & {} & \color {Red} {\text { Factor out a perfect square. }} \\ {x+4} & {=\pm 2 \sqrt{5}} & {} & \color {Red} {\text { Simplify: } \sqrt{4}=2}\end{array} \nonumber$

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

$x=-4 \pm 2 \sqrt{5} \quad \color {Red} \text { Subtract } 4 \text { from both sides. } \nonumber$

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

$x=-4-2 \sqrt{5} \quad \text { or } \quad x=-4+2 \sqrt{5} \nonumber$

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 $$\PageIndex{3}$$). Figure $$\PageIndex{3}$$: Checking $$-4-2 \sqrt{5}$$ and $$-4+2 \sqrt{5}$$ in the equation $$(x+4)^2 = 20$$.

Exercise $$\PageIndex{3}$$

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

$$-7+3 \sqrt{2},-7-3 \sqrt{2}$$

## Perfect Square Trinomials Revisited

Recall the squaring a binomial shortcut.

Squaring a binomial

If $$a$$ and $$b$$ are any real numbers, then: $(a±b)^2 = a^2 ±2ab + b^2 \nonumber$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:

\begin{aligned}(x+3)^{2} &=x^{2}+2(x)(3)+3^{2} \\ &=x^{2}+6 x+9 \end{aligned} \nonumber

\begin{aligned}(x-8)^{2} &=x^{2}-2(x)(8)+8^{2} \\ &=x^{2}-16 x+64 \end{aligned} \nonumber

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

$x^{2}+6 x+9=(x+3)^{2} \nonumber$

$x^{2}-16 x+64=(x-8)^{2} \nonumber$

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:$x^{2}-12 x+36=(x-6)^{2} \nonumber$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:$x^{2}+10 x+25=(x+5)^{2} \nonumber$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:$x^{2}-34 x+289=(x-17)^{2} \nonumber$Note that $$2(x)(17)=34 x$$, which is the middle term on the left. The solution checks.

Exercise $$\PageIndex{4}$$

Factor: $$x^2 + 30x + 225$$

$$(x+15)^{2}$$

## 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: $$\left(\dfrac{b}{2}\right)^{2}=\dfrac{b^{2}}{4}$$
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:

$x^{2}+b x+\dfrac{b^{2}}{4}=\left(x+\dfrac{b}{2}\right)^{2} \nonumber$

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:

$x^{2}+12 x+36=(x+6)^{2} \nonumber$

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.

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

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: $$\left(-\dfrac{3}{2}\right)^{2}=\dfrac{9}{4}$$
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:

$x^{2}-3 x+\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^{2} \nonumber$

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

Exercise $$\PageIndex{6}$$

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

$$x^{2}-5 x+\dfrac {10}{4}=\left(x-\dfrac {5}{2} \right)^{2}$$

## Solving Equations by Completing the Square

Consider the following nonlinear equation.

$x^2 =2x +2 \nonumber$

The standard approach is to make one side zero and factor.$x^2 −2x−2=0 \nonumber$ 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.$x^2 −2x =2 \nonumber$On the left, take one-half of the coeﬃcient of $$x: \left (\dfrac{1}{2} \right)(-2)=-1$$. Square the result: $$(-1)^{2}=1$$. Add this result to both sides of the equation.

$\begin{array}{l}{x^{2}-2 x+1=2+1} \\ {x^{2}-2 x+1=3}\end{array} \nonumber$

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

$(x-1)^{2}=3 \nonumber$

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.

$x-1=\pm \sqrt{3} \nonumber$

Finally, add $$1$$ to both sides of the equation.

$x=1 \pm \sqrt{3} \nonumber$

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 $$\PageIndex{4}$$). Figure $$\PageIndex{4}$$: Checking $$1-\sqrt{3}$$ and $$1+\sqrt{3}$$ in the equation $$x^2 =2x + 2$$.

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 $$x^2 =3−6x$$.

$$-3+2 \sqrt{3},-3-2 \sqrt{3}$$

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.

\begin{aligned} x^{2}-8 x-12=0 & \quad \color {Red} \text { Original equation. } \\ x^{2}-8 x=12 & \quad \color {Red} \text { Add } 12 \text { to both sides. } \end{aligned} \nonumber

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

\begin{aligned} x^{2}-8 x+16 & =12+16 \quad \color {Red} \text { Add } 16 \text { to both sides. } \\ (x-4)^{2} & =28 \quad \color {Red} \text { Factor left-hand side. } \\ x-4 &=\pm \sqrt{28} \quad \color {Red} \text { There are two square roots. }\end{aligned} \nonumber

$\begin{array}{rlrl}{x-4} & {=\pm \sqrt{4} \sqrt{7}} & {} & \color {Red} {\text { Factor out a perfect square. }} \\ {x-4} & {=\pm 2 \sqrt{7}} & {} & \color {Red} {\text { Simplify: } \sqrt{4}=2} \\ {x} & {=4 \pm 2 \sqrt{7}} & {} & \color {Red} {\text { Add } 4 \text { to both sides. }}\end{array} \nonumber$

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 $$\PageIndex{5}$$. Figure $$\PageIndex{5}$$: Drawing the graph of $$y = x^2 −8x−12$$.

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 $$\PageIndex{6}$$. Figure $$\PageIndex{6}$$: Calculating the zeros of $$y = x^2 −8x−12$$.

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 $$\PageIndex{7}$$). Figure $$\PageIndex{7}$$: Reporting your graphical solution on your homework.

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! Figure $$\PageIndex{8}$$: Approximating exact solutions $$x=4-2 \sqrt{7}$$ and $$x=4+2 \sqrt{7}$$.

Exercise $$\PageIndex{8}$$

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

$$-3-\sqrt{6},-3+\sqrt{6}$$ 