<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 10.4: Quadratic Equations by Completing the Square

Difficulty Level: 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
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:

(x±a)2=x2±2ax+a2\begin{align*}(x \pm a)^2 = x^2 \pm 2ax+a^2\end{align*}

• Point out that the leading coefficient is 1\begin{align*}1\end{align*}.
• In Example 3, a1\begin{align*}a \neq 1\end{align*}. Students learn to factor a\begin{align*}a\end{align*} from the whole expression before completing the square, whether or not the terms are multiples of a\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:

2x2+x3=0\begin{align*}2x^2+x-3=0\end{align*}

Solution: Although 2x2+x3=(2x+3)(x1)\begin{align*}2x^2+x-3=(2x+3)(x-1)\end{align*}, and therefore x=32\begin{align*}x= - \frac{3}{2}\end{align*} and x=1\begin{align*}x=1\end{align*}; we show that completing the square results in the same solutions.

Rewrite:

2x2+x=3\begin{align*}2x^2+x=3\end{align*}

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

x2+(12)x=32\begin{align*}x^2 + \left (\frac{1}{2} \right )x=\frac{3}{2}\end{align*}

Add the constant to both sides of the equation:

x2+(12)x+(14)2=32=(14)2\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.

(x+14)2(x+14)2(x+14)2=32+116=2416+116=2516\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:

(x+14)2x+14xxxx=2516=±54=14±54=1±54=1+54=1=154=32\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: x=32\begin{align*}x =- \frac{3}{2}\end{align*} and x=1\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 b\begin{align*}b\end{align*} coefficient. Remind students that dividing b\begin{align*}b\end{align*} by 2\begin{align*}2\end{align*} is equivalent to multiplying b\begin{align*}b\end{align*} by 1/2\begin{align*}1/2\end{align*}.

Solve the quadratic equation by completing the square:

3x2+5x3=0\begin{align*}3x^2 + 5x-3=0\end{align*}

Hint: Rewrite the equation:

3x2+5x=3.\begin{align*}3x^2 + 5x=3 .\end{align*}

And divide by the leading coefficient:

x2+53x=1\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 5/3\begin{align*}5/3\end{align*} by 1/2\begin{align*}1/2\end{align*} and then square:

(5312)2=(56)2\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 h\begin{align*}h\end{align*} and k\begin{align*}k\end{align*} correctly. See Examples 7-9.

Find the vertex of the parabola with equation:

y+3=2(x5)2\begin{align*}y+3= -2(x-5)^2\end{align*}

Solution: Rewrite the equation with minus signs:

y(3)=2(x5)2\begin{align*}y-(-3) = -2(x-5)^2\end{align*}

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

General Tip: Remind students that the leading coefficient must be 1\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.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: