# 10.4: Quadratic Equations by Completing the Square

**At Grade**Created by: CK-12

## 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.

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## Date Created:

Feb 22, 2012## Last Modified:

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