# Completing the Square (Part 2)

ID: pumom-bosap Illustrative Mathematics, CC BY 4.0
Subject: Algebra, Algebra 2
Standards: HSA-REI.AHSA-REI.B.4.b

17 questions

# Completing the Square (Part 2)

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##### Math Talk: Equations with Fractions

Solve each equation mentally.

1) ﻿$x + x= \frac{1}{4}$﻿

2) ﻿$(\frac{3}{2})^{2} = x$﻿

3) ﻿$\frac{3}{5} + x = \frac{9}{5}$﻿

4) ﻿$\frac{1}{12} + x = \frac{1}{4}$﻿

##### Solving Some Harder Equations

Solve these equations by completing the square.

5) ﻿$(x - 3)(x + 1) = 5$﻿

Show Work

6) ﻿$x^{2} + \frac{1}{2}x = \frac{3}{16}$﻿

Show Work

7) ﻿$x^{2} + 3x + \frac{8}{4} = 0$﻿

Show Work

8) ﻿$(7 - x)(3 - x) + 3 = 0$﻿

Show Work

9) ﻿$x^{2} + 1.6x + 0.63 = 0$﻿

Show Work
##### Spot Those Errors!

Here are four equations, followed by worked solutions of the equations. Each worked solution has at least one error.

Solve these equations by completing the square.

Then, look at the worked solution of the same equation as the one you solved. Find and describe the error or errors in the worked solution.

1. ﻿$x^2 + 14x = -24$﻿

2. ﻿$x^2 - 10x + 16 = 0$﻿

3. ﻿$x^2 + 2.4x = -0.8$﻿

4. ﻿$x^2 - \frac{6}{5}x + \frac{1}{5} = 0$﻿

Worked solutions (with errors):

Equation 1

﻿$\begin{array}{rcl} \\[-1em] x^2 + 14x &=& -24 \\[-1em] \\ \\[-1em] x^2 + 14x + 28 &=& 4 \\[-1em] \\ \\[-1em] (x + 7)^2 &=& 4 \\[-1em] \\ \\ \\[-1em] x + 7 = 2 &\text{or}& x + 7 = -2 \\[-1em] \\ \\[-1em] x = -5 &\text{or}& x = -9 \\[-1em] \end{array}$﻿

Equation 2

﻿$\begin{array}{rcl} \\[-1em] x^2 - 10x + 16 &=& 0 \\[-1em] \\ \\[-1em] x^2 - 10x + 25 &=& 9 \\[-1em] \\ \\[-1em] (x - 5)^2 &=& 9 \\[-1em] \\ \\ \\[-1em] x - 5 = 9 &\text{or}& x - 5 = -9 \\[-1em] \\ \\[-1em] x = 14 &\text{or}& x = -4 \\[-1em] \end{array}$﻿

Equation 3

﻿$\begin{array}{rcl} \\[-1em] x^2 + 2.4x &=& -0.8 \\[-1em] \\ \\[-1em] x^2 + 2.4x + 1.44 &=& 0.64 \\[-1em] \\ \\[-1em] (x + 1.2)^2 &=& 0.64 \\[-1em] \\ \\[-1em] x + 1.2 &=& 0.8 \\[-1em] \\ \\[-1em] x = -0.4 \\[-1em] \end{array}$﻿

Equation 4

﻿$\begin{array}{rcl} \\[-1em] x^2 - \frac{6}{5}x + \frac{1}{5} &=& 0 \\[-1em] \\ \\[-1em] x^2 - \frac{6}{5}x + \frac{9}{25} &=& \frac{9}{25} \\[-1em] \\ \\[-1em] \left(x - \frac{3}{5} \right)^2 &=& \frac{9}{25} \\[-1em] \\ \\ \\[-1em] x - \frac{3}{5} = \frac{3}{5} &\text{or}& x - \frac{3}{5} = -\frac{3}{5} \\[-1em] \\ \\[-1em] x = \frac{6}{5} &\text{or}& x = 0 \\[-1em] \end{array}$﻿

Equation 1: ﻿$x^{2} + 14x = -24$﻿

10) Solve the equation by completing the square.

Show Work

11) Find and describe the error(s) in the worked solution.

Equation 2: ﻿$x^{2} - 10x + 16 = 0$﻿

12) Solve the equation by completing the square.

Show Work

13) Find and describe the error(s) in the worked solution.

Equation 3: ﻿$x^{2} + 2.4x = -0.8$﻿

14) Solve the equation by completing the square.

Show Work

15) Find and describe the error(s) in the worked solution.

Equation 4: ﻿$x^{2} - \frac{6}{5}x + \frac{1}{5} = 0$﻿

16) Solve the equation by completing the square.

Show Work

17) Find and describe the error(s) in the worked solution.