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10.4: Quadratic Equations by Completing the Square

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

At the end of this lesson, students will be able to:

  • Complete the square of a quadratic expression.
  • Solve quadratic equations by completing the square.
  • Solve quadratic equations in standard form.
  • Graph quadratic equations in vertex form.
  • Solve real-world problems using functions by completing the square.

Vocabulary

Terms introduced in this lesson:

completing the square
perfect square trinomial
quadratic equations in standard form
quadratic equations in vertex form
parabola turns up, turns down

Teaching Strategies and Tips

Use Example 1 to introduce completing the square.

  • Remind students of binomial expansions to help them understand how a constant term turns the expression into a perfect square trinomial:

\begin{align*}(x \pm a)^2 = x^2 \pm 2ax+a^2\end{align*}

  • Point out that the leading coefficient is \begin{align*}1\end{align*}.
  • In Example 3, \begin{align*}a \neq 1\end{align*}. Students learn to factor \begin{align*}a\end{align*} from the whole expression before completing the square, whether or not the terms are multiples of \begin{align*}a\end{align*}. Complete the square of the resulting expression in parentheses.

Give completing the square geometrical meaning.

  • Use squares and rectangles for each term of the expression.
  • See paragraph preceding Example 4.
  • Have students make up a quadratic expression and ask them to complete the square in the geometrical interpretation as an assignment.

There are several reasons to have students learn completing the square:

  • It is used to derive the quadratic formula.
  • Quadratic equations can be rewritten in vertex form.
  • Equations of circles can be rewritten in graphing form.
  • Necessary in calculus.

Emphasize that completing the square finds roots

  • Regardless of whether the roots are integers, rational or irrational numbers.
  • Without having to list all the cases, unlike factoring.

Example:

Solve the following quadratic equation:

\begin{align*}2x^2+x-3=0\end{align*}

Solution: Although \begin{align*}2x^2+x-3=(2x+3)(x-1)\end{align*}, and therefore \begin{align*}x= - \frac{3}{2}\end{align*} and \begin{align*}x=1\end{align*}; we show that completing the square results in the same solutions.

Rewrite:

\begin{align*}2x^2+x=3\end{align*}

Divide by the leading coefficient; this results in an equation with \begin{align*}a = 1\end{align*}.

\begin{align*}x^2 + \left (\frac{1}{2} \right )x=\frac{3}{2}\end{align*}

Add the constant to both sides of the equation:

\begin{align*}x^2 + \left (\frac{1}{2} \right )x+ \left (\frac{1}{4} \right )^2 =\frac{3}{2} = \left (\frac{1}{4} \right )^2\end{align*}

Factor the perfect square trinomial and simplify.

\begin{align*}\left (x + \frac{1}{4} \right )^2 & = \frac{3}{2} + \frac{1}{16} \\ \left (x + \frac{1}{4} \right )^2 & = \frac{24}{16} + \frac{1}{16} \\ \left (x + \frac{1}{4} \right )^2 & = \frac{25}{16} \end{align*}

Take the square root of both sides:

\begin{align*}\sqrt{\left (x + \frac{1}{4} \right )^2} & =\sqrt{\frac{25}{16}} \\ x + \frac{1}{4} & = \pm \frac{5}{4} \\ x & = - \frac{1}{4} \pm \frac{5}{4} \\ x & = \frac{-1 \pm 5}{4} \\ \Rightarrow x & = \frac{-1 + 5}{4} = 1 \\ \Rightarrow x & = \frac{-1 - 5}{4} = - \frac{3}{2}\end{align*}

Answer: \begin{align*}x =- \frac{3}{2}\end{align*} and \begin{align*}x=1\end{align*} as expected.

Use Examples 7-9 to show how completing the square helps in graphing quadratic functions.

Error Troubleshooting

Dividing by the leading coefficient in Review Questions 7, 8, and 14 results in a fractional \begin{align*}b\end{align*} coefficient. Remind students that dividing \begin{align*}b\end{align*} by \begin{align*}2\end{align*} is equivalent to multiplying \begin{align*}b\end{align*} by \begin{align*}1/2\end{align*}.

Additional Example:

Solve the quadratic equation by completing the square:

\begin{align*}3x^2 + 5x-3=0\end{align*}

Hint: Rewrite the equation:

\begin{align*}3x^2 + 5x=3 .\end{align*}

And divide by the leading coefficient:

\begin{align*}x^2 + \frac{5}{3} x=1\end{align*}

To find the constant that must be added to both sides of the equation, multiply \begin{align*}5/3\end{align*} by \begin{align*}1/2\end{align*} and then square:

\begin{align*}\left (\frac{5}{3} \cdot \frac{1}{2} \right )^2 = \left (\frac{5}{6} \right )^2\end{align*}

General Tip: Remind students to rewrite the vertex form of a quadratic equation with minus signs to find \begin{align*}h\end{align*} and \begin{align*}k\end{align*} correctly. See Examples 7-9.

Additional Example:

Find the vertex of the parabola with equation:

\begin{align*}y+3= -2(x-5)^2\end{align*}

Solution: Rewrite the equation with minus signs:

\begin{align*}y-(-3) = -2(x-5)^2\end{align*}

The vertex is at \begin{align*}(h,k)\end{align*}, where \begin{align*}h = -3\end{align*} and \begin{align*}k = 5\end{align*}.

General Tip: Remind students that the leading coefficient must be \begin{align*}1\end{align*} before completing the square. They will need to divide or factor to accomplish this.

General Tip: Suggest that students rewrite equations in standard form before completing the square.

In Examples 10 and 11 and Review Questions 25 and 26, have students round in the last step.

Notes/Highlights Having trouble? Report an issue.

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