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Completing the Squares

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When dealing with algebraic expressions, it is always helpful to view them in their simplest form. That way, we can solve these expressions easily and determine possible patterns involved. In this case, we want to look at simplifying quadratic equations. So far, we have learned factoring methods such as grouping and identifying the greatest common factor. In this article, we shall be introduced to a new concept called completing the square. We will see the steps for solving quadratic equations by completing the square and examples of its application.

Expressions of the form${(x+a)}^{2}and{(x-a)}^{2}$are known as a **complete square**.

Expanding the first expression, we obtain

${(x+a)}^{2}=(x+a)(x+a)={x}^{2}+2ax+{a}^{2}$.

Similarly, the second expression becomes

${(x-a)}^{2}=(x-a)(x-a)={x}^{2}-2ax+{a}^{2}$,

upon expansion.

**Completing the square** is a method used to simplify a quadratic equation into an algebraic expression that can be easily solved. This technique can also be used to determine the maximum or minimum values of a quadratic equation and help with graphing.

The aim here is to convert the standard form of a quadratic equation so that it looks like the expression above. The general formula for completing the square is as follows.

For the quadratic equation, $a{x}^{2}+bx+c=0,$ we can complete the square of this expression by converting this into the form

$a{(x-h)}^{2}+k=0$ ,

where$h=\frac{b}{2a}andk=c-\frac{{b}^{2}}{4a}$. This form is known as the **vertex form** of a quadratic.

So what does it mean to complete the square? Before we get into some examples involving quadratic equations, it may be helpful to understand the geometry behind this method. Let us observe the diagram below.

In the first image, we have the red square and the green rectangle. Adding these two shapes together, we obtain the expression

${x}^{2}+bx$.

We want to rearrange this so that it looks like a square. Halving the width of the green rectangle, we obtain$\frac{{b}^{2}}{2}$.

Now rearranging these two new smaller green rectangles, we have the second image. Notice that we have a missing segment at the corner of the second image. Thus, to complete this square, we need to add the area of the blue square,${\left(\frac{b}{2}\right)}^{2}$. The complete square is shown in the third image. We can represent this algebraically as follows.

${x}^{2}+bx+{\left(\frac{b}{2}\right)}^{2}={\left(x+\frac{b}{2}\right)}^{2}$

where the term${\left(\frac{b}{2}\right)}^{2}$completes the square.

Here, we will deal with quadratic equations of the form x^{2} + bx + c, where the coefficient of the term x^{2} is 1.

From the previous section, it is important to note that we cannot add the term ${\left(\frac{b}{2}\right)}^{2}$without subtracting it from our initial expression. If we do not do this, our equation changes entirely. Let us show this with the following example.

Complete the square of the quadratic equation ${x}^{2}+4x+3=0$.

**Solution**

From the above equation, we have b = 4 and c = 3.

To complete the square, we have

${x}^{2}+4x\mathbf{+}{\mathbf{\left(}\frac{\mathbf{4}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}+3\mathbf{-}{\mathbf{\left(}\frac{\mathbf{4}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}=0$

Notice that we have to add ${\left(\frac{4}{2}\right)}^{2}$ and subtract ${\left(\frac{4}{2}\right)}^{2}$ to balance the equation. Now, observe that the first three terms above can be simplified into the form of a completed square as

${x}^{2}+4x+{\left(\frac{4}{2}\right)}^{2}={\left(x+\frac{4}{2}\right)}^{2}={(x+2)}^{2}$

Thus, the equation becomes

${x}^{2}+4x+3=0\Rightarrow {x}^{2}+4x+{\left(\frac{4}{2}\right)}^{2}+3-{\left(\frac{4}{2}\right)}^{2}=0\Rightarrow {(x+2)}^{2}-1=0$

As you can see, we've changed the equation from the standard form to vertex form with h = –2 and k = –1.

From the example above, observe that the variable x only appears once in the completed square expression. This allows us to solve the equation for x using basic algebra. Additionally, this means that we do not have to go through the hassle of performing the Quadratic Formula to solve a given quadratic equation. Let us return to the previous example.

Solve the equation${x}^{2}+4x+3=0$by completing the square.

**Solution**

Before, we found that

${x}^{2}+4x+3=0\Rightarrow {(x+2)}^{2}-1=0$

Solving for x, we obtain

${(x+2)}^{2}=1\Rightarrow x+2=\pm \sqrt{1}\Rightarrow x=\pm 1-2\Rightarrow x=-1-2andx=1-2$

Thus, we obtain two solutions $x=-3andx=-1$.

In some cases, the method above can be difficult to solve, especially when we are given quadratic equations with larger coefficients. It is always helpful to know a shortcut to make calculations faster. By completing the square, we want our expression to take the vertex form:

${(x-h)}^{2}+k$.Expanding the first term of this expression yields the definition of a complete square. The equation now becomes,

${(x-h)}^{2}+k={x}^{2}-2hx+{h}^{2}+k$.

The trick here is to "force" our given quadratic equation so that it takes the form of the expression above. Let us attempt this method in the previous example.

Complete the square of the following quadratic equation:

${x}^{2}+4x+3=0$.

**Solution **

Given the quadratic equation above, use the expression ${x}^{2}+2hx+{h}^{2}+k$to express the left-hand side as ${(x+h)}^{2}+k$.

Aligning these two expressions, we have

${x}^{2}+4x+3=0{x}^{2}+2hx+{h}^{2}+k=0$

From here we can see that 4x must take the form of 2hx, so d = 2 since $2\left(2x\right)=4x.$

The constant 3 must take the form h^{2} + k and since we know that d = 2,

Thus, we obtain

${x}^{2}+4x+3=0\Rightarrow {(x+2)}^{2}-1=0$

as we have solved before.

Before we move on to the next section, let us show another worked example.

Use the complete the square method to solve the quadratic equation ${x}^{2}-6x-12=0$.

**Solution**

From the above equation, we have b = –6 and c = –12.

We then note that

${x}^{2}-6x\mathbf{+}{\mathbf{\left(}\frac{\mathbf{6}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}-12\mathbf{-}{\mathbf{\left(}\frac{\mathbf{6}}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}=0$

Simplifying the first three terms as a complete square and solving, we obtain

${\left(x-\frac{6}{2}\right)}^{2}-21=0\Rightarrow {(x-3)}^{2}-21=0$

Now solving for x

${(x-3)}^{2}=21\Rightarrow x-3=\pm \sqrt{21}\Rightarrow x=\pm \sqrt{21}+3\Rightarrow x=-\sqrt{21}+3andx=\sqrt{21}+3$

Thus we have two solutions, correct to two decimals

$x=-\sqrt{21}+3\approx -1.58x=\sqrt{21}+3\approx 7.58$

In this section, we will deal with quadratic equations of the form ax^{2} + bx + c, where the coefficient of the term x^{2} is not equal to 1. For this form of quadratic equation, we can follow the steps below to complete the square.

**Step 1:** Given the standard form of the quadratic equation $a{x}^{2}+bx+c=0,$ divide all terms by a, that is the coefficient of x^{2}

${x}^{2}+\frac{b}{a}x+\frac{c}{a}=0$.

**Step 2: **Move the term $\frac{c}{a}$ to the right-hand side of the equation.

${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$.

**Step 3: **Complete the square on the left-hand side of the equation. Balance the equation by adding the same value to the right-hand side

${x}^{2}+\frac{b}{a}x\mathbf{+}{\mathbf{\left(}\frac{\mathbf{b}}{\mathbf{2}\mathbf{a}}\mathbf{\right)}}^{\mathbf{2}}=-\frac{c}{a}\mathbf{+}{\mathbf{\left(}\frac{\mathbf{b}}{\mathbf{2}\mathbf{a}}\mathbf{\right)}}^{\mathbf{2}}\Rightarrow {\left(x+\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+\frac{{\mathrm{b}}^{2}}{4{\mathrm{a}}^{2}}\Rightarrow {\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$

**Step 4: **Take the square root of both sides

$x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}=\pm \frac{\sqrt{{b}^{2}-4ac}}{\sqrt{4{a}^{2}}}\Rightarrow x+\frac{b}{2a}=\pm \frac{\sqrt{{b}^{2}-4ac}}{2a}$

**Step 5: **Solve to find x

$x=\pm \frac{\sqrt{{b}^{2}-4ac}}{2a}-\frac{b}{2a}\Rightarrow x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Notice that we have derived the Quadratic Formula using the method of completing the square!

Now going back to Step 3, we have deduced that

${\left(x+\frac{b}{2a}\right)}^{2}=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^{2}$

Bringing the terms from the right-hand side back to the left-hand side

${\left(x+\frac{b}{2a}\right)}^{2}+\frac{c}{a}-{\left(\frac{b}{2a}\right)}^{2}=0$

Multiplying the entire equation by a and simplifying, we obtain

$a{\left(x+\frac{b}{2a}\right)}^{2}+a\left(\frac{c}{a}\right)-a\left(\frac{{b}^{2}}{4{a}^{2}}\right)=0\Rightarrow a{\left(x+\frac{b}{2a}\right)}^{2}+c-\frac{{b}^{2}}{4a}=0$

Observe that the equation now takes the form $a{(x-h)}^{2}+k=0$ where

$h=\frac{b}{2a}andk=c-\frac{{b}^{2}}{4a}$

which is exactly the general form of completing the square for a quadratic equation as we have initially mentioned in the beginning. Below are some worked examples that demonstrate this.

Complete the square of $10{x}^{2}-2x+1=0$ and solve for x.

**Solution**

**Step 1:** Divide the expression by a = 10

${x}^{2}-\frac{2}{10}x+\frac{1}{10}=0\Rightarrow {x}^{2}-\frac{1}{5}x+\frac{1}{10}=0$

**Step 2:** Move the term $\frac{1}{10}$ to the other side

${x}^{2}-\frac{1}{5}x=-\frac{1}{10}$

Now multiplying the entire equation by a = 10, we obtain the vertex form

$10{\left(x-\frac{1}{10}\right)}^{2}+\frac{9}{10}=0$

**Step 4:** Taking the square root on both sides

$x-\frac{1}{10}=\pm \sqrt{-\frac{9}{100}}=\pm i\sqrt{\frac{9}{100}}=\pm \frac{3i}{10}$

**Note**: Remember that $\sqrt{-1}=i,\mathrm{sin}ce{i}^{2}=-1$

**Step 5:** Solving for x,

$x=\pm \frac{3i}{10}+\frac{1}{10}$

Thus, we have two solutions

$x=\frac{-3i+1}{10}andx=\frac{3i+1}{10}$

Complete the square of $-3{x}^{2}-4x+8=0$and solve for x.

**Solution **

**Step 1:** Divide the expression by a = –3

${x}^{2}-\frac{4}{(-3)}x+\frac{8}{(-3)}=0\Rightarrow {x}^{2}+\frac{4}{3}x-\frac{8}{3}=0$

**Step 2:** Move the term $-\frac{8}{3}$ to the other side

${x}^{2}+\frac{4}{3}x=\frac{8}{3}$

**Step 3:** Complete the square and balance the equation

${x}^{2}+\frac{4}{3}x+{\left(\frac{4}{2\left(3\right)}\right)}^{2}=\frac{8}{3}+{\left(\frac{4}{2\left(3\right)}\right)}^{2}\Rightarrow {x}^{2}+\frac{4}{3}x+{\left(\frac{2}{3}\right)}^{2}=\frac{8}{3}+{\left(\frac{2}{3}\right)}^{2}\Rightarrow {\left(x+\frac{2}{3}\right)}^{2}=\frac{28}{9}\Rightarrow {\left(x+\frac{2}{3}\right)}^{2}-\frac{28}{9}=0$

Now multiplying the entire equation by a = –3, we obtain the vertex form

$-3{\left(x+\frac{2}{3}\right)}^{2}+\frac{28}{3}=0$

**Step 4:** Taking the square root on both sides

$x+\frac{2}{3}=\pm \sqrt{\frac{28}{9}}=\pm \frac{2\sqrt{7}}{3}$

**Step 5:** Solving for x,

$x=\pm \frac{2\sqrt{7}}{3}-\frac{2}{3}=\frac{\pm 2\sqrt{7}-2}{3}$

Therefore, we have two solutions

$x=\frac{-2(\sqrt{7}+1)}{3}and\frac{2(\sqrt{7}-1)}{3}$

Completing the square also helps us determine the maximum and minimum values of a given quadratic equation. By doing so, we can locate this value and plot the graph of a quadratic equation more accurately.

The **vertex **is a point at which the curve on a graph turns from decreasing to increasing or from increasing to decreasing. This is also known as a turning point.

The **maximum value** is the highest point of the curve in a graph. This is also known as the maximum turning point or local maxima.

The **minimum value** is the lowest point of the curve in a graph. This is also known as the minimum turning point or local minima.

For the general form of a quadratic equation, the maximum and minimum values on a graph take on the following two conditions.

Essentially, if the coefficient of x^{2 }is positive, then the graph curves downwards and if the coefficient of x^{2 }is negative, then the graph curves upwards. From the general formula of completing the square, when the coefficient of x^{2} is 1,

${(x-h)}^{2}+k=0$,

the x and y coordinates of the turning point, or the vertex, can be found by the point (h, k). Similarly, when the coefficient of x^{2} is not 1,

$a{(x-h)}^{2}+k=0$,

the x and y coordinates of the turning point, or the vertex, can be found by the same point, (h, k). Note that the value of a does not affect the position of the vertex!

Let us look for the maximum and minimum values for the last two examples from the previous section.

Determine whether the quadratic equation $10{x}^{2}-2x+1=0$ has a maximum or minimum value. Hence, find the coordinates of its turning point.

**Solution **

The coefficient of the term x^{2} is positive, as a = 10. Thus, we have a minimum value. In this case, the curve opens up. From the derivation of the completed square form of this expression, we obtain

$10{\left(x-\frac{1}{10}\right)}^{2}+\frac{9}{10}=0$.

Here, $x=\frac{1}{10}$.

**Remember that the value of a does not vary the x-value of the vertex!**

Thus, the minimum value is $\frac{9}{10}$ when $x=\frac{1}{10}$.

The coordinates of the minimum turning point is $\left(\frac{1}{10},\frac{9}{10}\right).$ The graph is shown below.

Determine whether the quadratic equation $-3{x}^{2}-4x+8=0$ has a maximum or minimum value. Hence, find the coordinates of its turning point.

**Solution **

The coefficient of the term x^{2} is negative, as a = –3. Thus, we have a maximum value. In this case, the curve opens down. From the derivation of the completed square form of this expression, we obtain

$-3{\left(x+\frac{2}{3}\right)}^{2}+\frac{28}{3}=0$.

Here, $x=-\frac{2}{3}$.

Thus, the maximum value is $\frac{28}{3}$ when $x=-\frac{2}{3}$.

The coordinates of the maximum turning point is $\left(-\frac{2}{3},\frac{28}{3}\right).$ The graph is shown below.

- For the quadratic equation $a{x}^{2}+bx+c=0,$ the standard form of the completed square form is $a{(x-h)}^{2}+k=0$ where $h=-\frac{b}{2a}andh=c-\frac{{b}^{2}}{4a}$
- Completing the square is a method used to
- Simplify a quadratic equation
- Determine the maximum or minimum values

- To complete the square and solve the quadratic equation, we must
- Divide the expression by the coefficient of x
^{2} - Move the third term to the right-hand side
- Complete the square and balance the equation
- Take the square roots of both sides
- Solve for x

- Divide the expression by the coefficient of x
- If the coefficient of x
^{2 }is positive, then we have a minimum value. - If the coefficient of x
^{2 }is negative, then we have a maximum value. - The coordinates for the turning point is (h, k).

To solve a quadratic equation by completing the square, we must

- Divide the expression by the coefficient of
*x*^{2} - Move the third term to the right-hand side
- Complete the square and balance the equation
- Take the square roots of both sides
- Solve for
*x*

Yes, all quadratic equations can be solved by completing the square** **

The formula for completing the square is *a(x–d) ^{2}+e=0, where d=–b/2a and e=c-b^{2}/4a*

To solve a quadratic equation by completing the square, we must

- Divide the expression by the coefficient of
*x*^{2} - Move the third term to the right-hand side
- Complete the square and balance the equation
- Take the square roots of both sides
- Solve for
*x*

More about Completing the Squares

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