Lesson 14

Completing the Square (Part 3)

  • Let’s complete the square for some more complicated expressions.

14.1: Perfect Squares in Two Forms

Elena says, “\((x+3)^2\) can be expanded into \(x^2 + 6x + 9\). Likewise, \((2x+3)^2\) can be expanded into \(4x^2 + 6x + 9\).”

Find an error in Elena’s statement and correct the error. Show your reasoning.

14.2: Perfect in A Different Way

  1. Write each expression in standard form:

    1. \((4x+1)^2\)
    2. \((5x-2)^2\)
    3. \((\frac12 x + 7)^2\)
    4. \((3x+n)^2\)
    5. \((kx+m)^2\)
  2. Decide if each expression is a perfect square. If so, write an equivalent expression of the form \((kx+m)^2\). If not, suggest one change to turn it into a perfect square.

    1. \(4x^2 + 12x + 9\)
    2. \(4x^2 + 8x+ 25\)

14.3: When All the Stars Align

  1. Find the value of \(c\) to make each expression in the left column a perfect square in standard form. Then, write an equivalent expression in the form of squared factors. In the last row, write your own pair of equivalent expressions.
    standard form \((ax^2+bx+c)\) squared factors \((kx+m)^2\)
    \(100x^2+80x+c\)  
    \(36x^2-60x+c\)  
    \(25x^2+40x+c\)  
    \(0.25x^2-14x+c\)  
                                       
  2. Solve each equation by completing the square:

    \(25x^2 + 40x = \text-12\)

    \(36x^2 - 60x + 10 = \text-6\)

14.4: Putting Stars into Alignment

Here are three methods for solving \(3x^2 + 8x + 5 = 0\).

Try to make sense of each method.

Method 1:

\(\displaystyle \begin {align}3x^2 + 8x + 5 &= 0\\ (3x + 5)(x + 1) &= 0 \end{align}\)

\(\displaystyle \begin {align} x = \text- \frac53 \quad \text{or} \quad x = \text-1\end {align}\)

Method 2:

\(\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\ (3x)^2 + 8(3x) + 15 &= 0\\ U^2 + 8U + 15 &= 0\\ (U+5)(U+3) &= 0 \end{align}\)
\(\displaystyle \begin {align} U = \text-5 \quad &\text{or} \quad U = \text-3\\3x = \text-5 \quad &\text{or} \quad 3x = \text-3\\ x = \text- \frac53 \quad &\text{or} \quad x = \text-1 \end{align}\)

Method 3:

\(\displaystyle \begin {align} 3x^2 + 8x + 5 &= 0\\ 9x^2 + 24x + 15 &= 0\\9x^2 + 24x + 16 &= 1\\(3x + 4)^2 &= 1 \end{align}\)

\(\displaystyle \begin {align} 3x+4 = 1 \quad & \text{or} \quad 3x+4 = \text-1\\x = \text-1 \quad & \text{or} \quad x = \text- \frac53 \end {align}\)

Once you understand the methods, use each method at least one time to solve these equations.

  1. \(5x^2 + 17x + 6 = 0\)
  2. \(6x^2 + 19x = \text-10\)
  3. \(8x^2 - 33x + 4 = 0\)
  4. \(8x^2 - 26x = \text-21\)
  5. \(10x^2 + 37x = 36\)
  6. \(12x^2 + 20x - 77=0\)


Find the solutions to \(3x^2 -6x + \frac{9}{4} = 0\). Explain your reasoning.

Summary

In earlier lessons, we worked with perfect squares such as \((x+1)^2\) and \((x-5)(x-5)\). We learned that their equivalent expressions in standard form follow a predictable pattern:

  • In general, \((x+m)^2\) can be written as \(x^2 + 2mx + m^2\).
  • If a quadratic expression is of the form \(ax^2 + bx + c\), and the value of \(a\) is 1, then the value of \(b\) is \(2m\), and the value of \(c\) is \(m^2\) for some value of \(m\).

In this lesson, the variable in the factors being squared had a coefficient other than 1, for example \((3x+1)^2\) and \((2x-5)(2x-5)\). Their equivalent expression in standard form also followed the same pattern we saw earlier.

squared factors standard form
\((3x+1)^2\) \((3x)^2 + 2(3x)(1) + 1^2 \quad \text{or} \quad 9x^2 +6x + 1\)
\((2x-5)^2\) \((2x)^2 + 2(2x)(\text-5) + (\text-5)^2 \quad \text{or} \quad 4x^2 -20x + 25\)

In general, \((kx+m)^2\) can be written as:

\(\displaystyle (kx)^2 + 2 (kx)(m) + m^2\)

or

\(\displaystyle k^2 x^2 + 2kmx + m^2\)

If a quadratic expression is of the form \(ax^2 + bx + c\), then:

  • the value of \(a\) is \(k^2\)
  • the value of \(b\) is \(2km\)
  • the value of \(c\) is \(m^2\)

We can use this pattern to help us complete the square and solve equations when the squared term \(x^2\) has a coefficient other than 1—for example: \(16x^2 + 40x = 11\).

What constant term \(c\) can we add to make the expression on the left of the equal sign a perfect square? And how do we write this expression as squared factors?

  • 16 is \(4^2\), so the squared factors could be \((4x + m)^2\).
  • 40 is equal to \(2(4m)\), so \(2(4m) = 40\) or \(8m=40\). This means that \(m = 5\).
  • If \(c\) is \(m^2\), then \(c = 5^2\) or \(c=25\).
  • So the expression \(16x^2 + 40x + 25\) is a perfect square and is equivalent to \((4x+5)^2\).

Let’s solve the equation \(16x^2 + 40x = 11\) by completing the square!

\(\displaystyle \begin {align} 16x^2 + 40x &= 11\\ 16x^2 + 40x + 25 &= 11 + 25\\ (4x + 5)^2 &=36\\\\4x+5 = 6 \quad &\text {or} \quad 4x+5= \text-6\\ 4x = 1 \quad &\text {or} \quad 4x = \text-11\\ x=\frac14 \quad &\text {or} \quad x = \text- \frac{11}{4} \end {align}\).

Glossary Entries

  • completing the square

    Completing the square in a quadratic expression means transforming it into the form \(a(x+p)^2-q\), where \(a\), \(p\), and \(q\) are constants.

    Completing the square in a quadratic equation means transforming into the form \(a(x+p)^2=q\).

  • perfect square

    A perfect square is an expression that is something times itself. Usually we are interested in situations where the something is a rational number or an expression with rational coefficients.

  • rational number

    A rational number is a fraction or the opposite of a fraction. Remember that a fraction is a point on the number line that you get by dividing the unit interval into \(b\) equal parts and finding the point that is \(a\) of them from 0. We can always write a fraction in the form \(\frac{a}{b}\) where \(a\) and \(b\) are whole numbers, with \(b\) not equal to 0, but there are other ways to write them. For example, 0.7 is a fraction because it is the point on the number line you get by dividing the unit interval into 10 equal parts and finding the point that is 7 of those parts away from 0. We can also write this number as \(\frac{7}{10}\).

    The numbers \(3\), \(\text-\frac34\), and \(6.7\) are all rational numbers. The numbers \(\pi\) and \(\text-\sqrt{2}\) are not rational numbers, because they cannot be written as fractions or their opposites.