Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

Activity

Access the Desmos guide PDF for tips on solving problems with the Desmos graphing calculator.

Here is an example of an equation being solved by graphing and by completing the square.

Graph of a parabola on a coordinate plane. Two points (negative 4.414, 0) and (negative 1.586, 0) have been labeled on the parabola. The x-axis has a scale of 1 extending from negative 8 to 4. The y-axis extends from negative 4 to 4 with a scale of 1.

$\begin{array}{rcl} x^{2} + 6 x + 7 & = & 0 \\ x^{2} + 6 x + 9 & = & 2 \\ (x + 3)^{2} & = & 2 \\ x + 3 & = & \pm \sqrt{2} \\ x & = & - 3 \pm \sqrt{2} \end{array}$

Verify: $\sqrt{2}$ is approximately 1.414. So $- 3 + \sqrt{2} \approx - 1.586$ and $- 3 - \sqrt{2} \approx - 4.414$ .

For each equation, find the exact solutions by completing the square and the approximate solutions by graphing. Then, verify that the solutions found using the two methods are close. If you get stuck, refer back to the example.

1.

1. $x^{2} + 4 x + 1 = 0$

a. Find the exact solutions by completing the square.

Compare your answer:

$x^{2} + 4 x + 1 = 0 x^{2} + 4 x = - 1 x^{2} + 4 x + 4 = - 1 + 4 x^{2} + 4 x + 4 = 3 (x + 2)^{2} = 3 \sqrt{(x + 2)^{2}} = \sqrt{3} x + 2 = \pm \sqrt{3} x = - 2 \pm \sqrt{3}$

b. Graph the equation to find approximate solutions.

Use the Desmos graphing tool or technology outside the course.

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2.

Compare your answer:

A graph of a parabola opening upwards crossing the x-axis at approximately (negative 3.732, 0) and (negative 0.268, 0). Both the x- and y-axes extend from negative 5 to 5 with a scale of 1.

3.

c. Are the solutions you found using the two methods close?

Yes. Solutions were the same: $x \approx - 0.268$ and $x \approx - 3.732$ .

4.

2. $x^{2} - 10 x + 18 = 0$

a. Find the exact solutions by completing the square.

Compare your answer:

$x^{2} - 10 x + 18 = 0 x^{2} - 10 x = - 18 x^{2} - 10 x + 25 = - 18 + 25 x^{2} - 10 x + 25 = 7 (x - 5)^{2} = 7 \sqrt{(x - 5)^{2}} = \sqrt{7} x - 5 = \pm \sqrt{7} x = 5 \pm \sqrt{7}$

b. Graph the equation to find approximate solutions.

Use the Desmos graphing tool or technology outside the course.

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5.

Compare your answer:

A graph of a parabola opening upwards, crossing the x-axis at points (2.354, 0) and (7.646, 0). The x-axis extends from negative 1 to approximately 18 with a scale of 1. The y-axis extends from negative 12 to 7 with a scale of 1.

6.

c. Are the solutions you found using the two methods close?

Yes. Solutions were the same: $x \approx 7.646$ and $x \approx 2.354$ .

7.

3. $x^{2} + 5 x + \frac{1}{4} = 0$

a. Find the exact solutions by completing the square.

Compare your answer:

$\begin{matrix} x^{2} + 5 x + \frac{1}{4} & = & 0 \\ x^{2} + 5 x & = & - \frac{1}{4} \\ x^{2} + 5 x + {(\frac{5}{2})}^{2} & = & - \frac{1}{4} + {(\frac{5}{2})}^{2} \\ x^{2} + 5 x + \frac{25}{4} & = & - \frac{1}{4} + \frac{25}{4} \\ x^{2} + 5 x + \frac{25}{4} & = & \frac{24}{4} \\ x^{2} + 5 x + \frac{25}{4} & = & 6 \\ {(x + \frac{5}{2})}^{2} & = & 6 \\ \sqrt{{(x + \frac{5}{2})}^{2}} & = & \sqrt{6} \\ x + \frac{5}{2} & = & \pm \sqrt{6} \\ x & = & - \frac{5}{2} \pm \sqrt{6} \end{matrix}$

b. Graph the equation to find approximate solutions.

Use the Desmos graphing tool or technology outside the course.

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8.

Compare your answer:

Graph of a parabola on a coordinate grid. The points ( negative 4.949, 0) and (negative 0.051, 0) are labeled on the parabola. The x-axis extends from negative 8 to approximately 8 with a scale of 1. The y-axis extends from negative 10 to 6 with a scale of 1.

9.

c. Are the solutions you found using the two methods close?

Yes. Solutions were the same: $x \approx - 4.949$ and $x \approx - 0.051$ .

10.

4. $x^{2} + \frac{8}{3} x + \frac{14}{9} = 0$

a. Find the exact solutions by completing the square.

Compare your answer:

$\begin{matrix} x^{2} + \frac{8}{3} x + \frac{14}{9} & = & 0 \\ x^{2} + \frac{8}{3} x & = & - \frac{14}{9} \\ x^{2} + \frac{8}{3} x + {(\frac{1}{2} \cdot \frac{8}{3})}^{2} & = & - \frac{14}{9} + {(\frac{1}{2} \cdot \frac{8}{3})}^{2} \\ x^{2} + \frac{8}{3} x + {(\frac{4}{3})}^{2} & = & - \frac{14}{9} + {(\frac{4}{3})}^{2} \\ x^{2} + \frac{8}{3} x + \frac{16}{9} & = & - \frac{14}{9} + \frac{16}{9} \\ x^{2} + \frac{8}{3} x + \frac{16}{9} & = & \frac{2}{9} \\ {(x + \frac{4}{3})}^{2} & = & \frac{2}{9} \\ \sqrt{{(x + \frac{4}{3})}^{2}} & = & \sqrt{\frac{2}{9}} \\ x + \frac{4}{3} & = & \pm \frac{\sqrt{2}}{3} \\ x & = & - \frac{4}{3} \pm \frac{\sqrt{2}}{3} \end{matrix}$

b. Graph the equation to find approximate solutions.

Use the Desmos graphing tool or technology outside the course.

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11.

Compare your answer:

Graph of a parabola on a coordinate grid. The points (negative 1.805, 0) and (negative 0.862, 0) are labeled on the parabola. The x-axis extends from negative 2.6 to approximately 1.2 with a scale of 0.2. The y-axis extends from negative 1.6 to 2.2 with a scale of 0.2.

12.

c. Are the solutions you found using the two methods close?

Yes. Solutions were the same: $x \approx - 0.862$ and $x \approx - 1.805$ .

Are you ready for more?

Extending Your Thinking

1.

What quadratic equation of the form $a x^{2} + b x + c = 0$ would have the solutions $x = 5 - \sqrt{2}$ and $x = 5 + \sqrt{2}$ ?

Enter the quadratic equation.

Compare your answer:

$x^{2} - 10 x + 23 = 0$ . Working backward, these are solutions to $(x - 5)^{2} = 2$ . In standard form, this is $x^{2} - 10 x + 23 = 0$ .

Self Check

Which of the following are the solutions to

(x-2)^2=12

, rounded to the nearest thousandth?

$x \approx 3.162$
$x=14$
$x \approx -1.464, x \approx 5.464$
$x \approx -3.162, x \approx 3.162$

Additional Resources

Using the Calculator for Approximate Solutions

Let's look at the solutions to the equations from 9.5.2: Additional Resources:

A more compact way to write the two solutions to the equation $(x + 4)^{2} = 11$ is: $x = - 4 \pm \sqrt{11}$ .

How large or small are those numbers? Are they positive or negative? We can use a calculator to compute the approximate values of both expressions:

$- 4 + \sqrt{11} \approx - 0.683$ or $- 4 - \sqrt{11} \approx - 7.317$

We can also approximate the solutions by graphing. The equation $(x + 4)^{2} = 11$ is equivalent to $(x + 4)^{2} - 11 = 0$ , so we can graph the function $y = (x + 4)^{2} - 11$ and find its zeros by locating the $x$ -intercepts of the graph.

Graph of a parabola on a coordinate grid. The points (negative 7.317, 0) and (negative 0.683, 0) are labeled on the parabola. The x-axis extends from negative 14 to 6 with a scale of 2. The y-axis extends from negative 14 to 4 with a scale of 2.

Try it

Try It: Using the Calculator for Approximate Solutions

Approximate the solutions to $x = 3 \pm \sqrt{19}$ .

Here is how to approximate the solutions:

Step 1 - Write two expressions.

$x = 3 + \sqrt{19}, x = 3 - \sqrt{19}$

Step 2 - Use the calculator to approximate the solutions to three decimal places.

$x \approx - 1.359, x \approx 7.359$

9.5.3 Finding Irrational Solutions by Completing the Square

Activity

Are you ready for more?

Extending Your Thinking

Additional Resources

Using the Calculator for Approximate Solutions

Try It: Using the Calculator for Approximate Solutions