Completing the Square

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.

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Team Completing the Square Teachers

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    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.

    What is "completing the square"?

    If a given quadratic equation can be factored to a perfect square of a linear binomial, it can be solved easily by equating the resulting binomial to 0 and solving it. For example, if we factor a quadratic equation to yield

    \[(ax + b)^2 = 0\]

    then we can proceed to the final solution as follows:

    \[ax + b = 0 \Rightarrow ax = -b \Rightarrow x = -\frac{b}{a}\]

    However, it's difficult to directly reduce many quadratic equations to a perfect square. For these quadratics, we use a method called completing the square.

    Using the completing the square method, we try to obtain a perfect square trinomial on the left-hand side of the equation. We then proceed to solve the equation using the square roots.

    Using the completing the square method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equation.

    In other words, completed squares are expressions of the form \((x+a)^2\) and \((x-a)^2\).

    Completing the square formula

    In this article, we will go through the more formal steps of the completing the square method. But first, in this section, we look at a bit of a cheat sheet for solving quadratic equations by completing the square.

    Given a quadratic equation of the form,

    \(ax^2 + bx+c = 0\)

    we convert it into

    \((x+d)^2 = e \text{, where } d = \frac{b}{2a} \text{ and } e = \frac{b^2}{4a^2}- \frac{c}{a}\). This form is known as the vertex form of a quadratic.

    Directly implementing this formula will also give you the answer.

    Completing the square method

    While you can directly use the formula stated above, there is a more deliberate step-by-step method to solving quadratic equations using the completing the square method.

    Note that in exams you would need to solve using the step-by-step method, so it is a good idea to get familiar with the process.

    If you are given a quadratic equation of the form \(ax^2 + bx + c = 0\), follow the steps below to solve it using the completing the square method:

    1. If a (coefficient of x2) is not 1, divide each term by a.

      This yields an equation of the form \(x^2 + \frac{b}{a} x + \frac{c}{a} = 0\)

    2. Move the constant term (\(\frac{c}{a}\)) to the right-hand side.

      This yields an equation of the form \(x^2 + \frac{b}{a} x = -\frac{c}{a}\)

    3. Add the appropriate term to complete the square of the left-hand side of the equation. Do the same addition on the right-hand side to keep the equation balanced.

      Hint: the appropriate term should be equal to \((\frac{b}{2a})^2\).

      The equation should now be in the form \((x+d)^2 = e\)

    4. Now that you have a perfect square on the left-hand side, you can find the roots of the equation by taking square roots.

    Let us take a look at some examples to illustrate this.

    Geometrical representation of completing the square

    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.

    Graphic representation of completing the squareFig. 1. Graphic representation of completing the square.

    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, \((\frac{b}{2})^2\). The complete square is shown in the third image. We can represent this algebraically as follows.

    \[x^2+bx +(\frac{b}{2})^2 = (x+\frac{b}{2})^2\]

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

    Completing the square examples

    Here are a few examples with solutions for completing the squares.

    Solve for x : \(2x^2 + 8x+3 = 0\)

    Solution:

    Step 1 – Divide each term by 2:

    \(x^2 + 4x + \frac{3}{2} = 0\)

    Step 2 –Move the constant term to the right-hand side.

    \(x^2 + 4x = -\frac{3}{2}\)

    Step 3 –Complete the square by adding 4 to both sides.

    \(x^2 + 4x + 4 = -\frac{3}{2} + 4 \Rightarrow (x+2)^2 = \frac{5}{2}\)

    Step 4 –Find the roots by taking square roots.

    \(x+2 = \pm\sqrt{\frac{5}{2}} \Rightarrow x = -2 \pm \sqrt{\frac{5}{2}}\)

    Thus, the roots of the equation are

    \(x = -2 + \sqrt{\frac{5}{2}} \text{ and } x = -2 - \sqrt{\frac{5}{2}} \)

    Solve for x : \(x^2-6x-7 = 0\)

    Solution:

    Step 1 – The coefficient of x2 is 1. So we can move on to step 2.

    Step 2 – Move the constant term to the right-hand side.

    \(x^2-6x = 7\)

    Step 3 – Complete the square by adding 9 to both sides.

    \(x^2 -6x +9 = 7 + 9 \Rightarrow (x-3)^2 = 16\)

    Step 4 – Find the roots by taking square roots.

    \(x-3 = \pm \sqrt{16} \Rightarrow x= 3 \pm 4\)

    Thus, the roots of the equation are

    \(x = 3+4 = 7 \text{ and } x= 3-4 = -1\)

    Remember the formula we discussed earlier in the article. Let us now try solving the above example directly using the completing the squares formula.

    Do keep in mind that during your exam, you should use the method described above instead of directly inserting values into the formula.

    Solve for x: \(x^2-6x-7 = 0\)

    Solution:

    Let us directly put the equation in the form

    \((x+d)^2 = e \text{, where } d = \frac{b}{2a} \text{ and } e = \frac{b^2}{4a^2} - \frac{c}{a}.

    From the equation: a = 1, b = -6, c = -7. So:

    \(d = \frac{-6}{2 \cdot 1} = -3e = \frac{-6^2}{4 \cdot 1^2} - \frac{-7}{1} = 9+7 = 16\)

    This gives us

    \((x+d)^2 = e \Rightarrow (x-3)^2 = 16\)

    which is exactly what we got using the method in the previous example. From here on, you can follow the process in the same way as in the above example to obtain the roots, 7 and -1.

    While you should not solve questions like this in a written examination, this can be a very useful short cut if you need to rapidly find the roots of a quadratic equation or if you want to cross-check whether the answer you have found using the former method is accurate.

    Identifying the Maximum and Minimum Values of a Quadratic Equation

    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.

    Maximum and minimum values of a plot, completing the square, studysmarterFig. 2. A general plot of the maximum and minimum values of a quadratic equation.

    Essentially, if the coefficient of x2 is positive, then the graph curves downwards and if the coefficient of x2 is negative, then the graph curves upwards. From the general formula of completing the square, when the coefficient of x2 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 x2 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 \(10x^2 -2x +1\) has a maximum or minimum value. Hence, find the coordinates of its turning point.

    Solution

    The coefficient of the term x2 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

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

    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 \(\frac{1}{10}\).

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

    Graph with a minimum, studysmarterFig. 3. Problem graph #1.

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

    Solution

    The coefficient of the term x2 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

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

    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 \((-\frac{2}{3}, \frac{28}{3})\) The graph is shown below.

    Graph with a maximum, studysmarterFig. 4. Problem graph #2.

    Completing the Square - Key takeaways

    • Many quadratic equations are very difficult to directly reduce to a perfect square. For such quadratics, we can use the method called completing the square.
    • Using the completing the square method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equation.
    • Using the completing the square method we transform a quadratic equation of the form\(ax^2 + bx + c = 0\) into \((x+d)^2 = e \text{,where } d= \frac{b}{2a} \text{ and } e = \frac{b^2}{4a^2} - \frac{c}{a}\)
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    Completing the Square
    Frequently Asked Questions about Completing the Square

    What is the completing the square method?

    Using the completing the square method, we add or subtract terms to both sides of a quadratic equation until we have a perfect square trinomial on one side of the equation.

    What is the formula of completing the square?

    Using the completing the square method we transform a quadratic equation of the form ax²+bx+c=0 into (x+d)²=e, where d=b/2a and e=b²/4a² - c/a

    What are the steps of completing the square?

    If you are given a quadratic equation of the form ax²+bx+c=0, follow the steps below to solve it using the completing the square method:


    1. If a (coefficient of x2) is not 1, divide each term by a. 
    2. Move the constant term to the right hand side.
    3. Add the appropriate term to complete the square of the left hand side of the equation. Do the same addition on the right hand side to keep the equation balanced.
    4. Now that you have a perfect square on the left hand side, you can find the roots of the equation by taking square roots. 

    What is an example of completing the square method?

    Beolow is an example of completing the squares:

    Solve for x : Solution

    Step 1 – Divide each term by 2.


    Step 2 –Move the constant term to the right-hand side.


    Step 3 –Complete the square by adding 4 to both sides.


    Step 4 –Find the roots by taking square roots.


    Thus, the roots of the equation are


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