# Completing the Square (Part 2)

ID: juriz-fijin Illustrative Mathematics, CC BY 4.0
Subject: Algebra, Algebra 2

26 questions

# Completing the Square (Part 2)

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##### Problem 1

Add the number that would make the expression a perfect square. Next, write an equivalent expression in factored form.

1) Add a number to ﻿$x^{2} + 3x$﻿ in order to write it as a perfect square.

Show Work

2) Write an equivalent expression in factored form.

Show Work

3) Add a number to ﻿$x^{2} + 0.6x$﻿ in order to write it as a perfect square.

Show Work

4) Write an equivalent expression in factored form.

Show Work

5) Add a number to ﻿$x^{2} - 11x$﻿ in order to write it as a perfect square.

Show Work

6) Write an equivalent expression in factored form.

Show Work

7) Add a number to ﻿$x^{2} - \frac{5}{2}x$﻿ in order to write it as a perfect square.

Show Work

8) Write an equivalent expression in factored form.

Show Work

9) Add a number to ﻿$x^{2} + x$﻿ in order to write it as a perfect square.

Show Work

10) Write an equivalent expression in factored form.

Show Work
##### Problem 2

11) Noah is solving the equation ﻿$x^2 + 8x + 15 = 3$﻿. He begins by rewriting the expression on the left in factored form and writes ﻿$(x + 3)(x + 5) = 3$﻿. He does not know what to do next.

Noah knows that the solutions are ﻿$x = -2$﻿ and ﻿$x = -6$﻿, but is not sure how to get to these values from his equation.

Solve the original equation by completing the square.

##### Problem 3

An equation and its solutions are given. Explain or show how to solve the equation by completing the square.

12) ﻿$x^{2} + 20x + 50 = 14$﻿. The solutions are ﻿$x = -18$﻿ and ﻿$x = -2$﻿.

13) ﻿$x^{2} + 1.6x = 0.36$﻿. The solutions are ﻿$x = -1.8$﻿ and ﻿$x = 0.2$﻿.

14) ﻿$x^{2} - 5x = \frac{11}{4}$﻿. The solutions are ﻿$x = \frac{11}{2}$﻿ and ﻿$x = \frac{-1}{2}$﻿.

##### Problem 4

Solve each equation.

15) ﻿$x^{2} - 0.5x = 0.5$﻿

Show Work

16) ﻿$x^{2} + 0.8x = 0.09$﻿

Show Work

17) ﻿$x^{2} + \frac{13}{3}x = \frac{56}{36}$﻿

Show Work
##### Problem 5

Match each quadratic expression given in factored form with an equivalent expression in standard form. One expression in standard form has no match.

18) ﻿$(2 \ + \ x)(2 \ - \ x)$﻿

a)

﻿$x^{2}$﻿ - 4

b)

81 - ﻿$x^{2}$﻿

c)

﻿$x^{2}$﻿ - ﻿$y^{2}$﻿

d)

4 - ﻿$x^{2}$﻿

e)

﻿$x^{2}$﻿ - 81

19) ﻿$(x \ + \ 9)(x \ - \ 9)$﻿

a)

﻿$x^{2}$﻿ - 4

b)

81 - ﻿$x^{2}$﻿

c)

﻿$x^{2}$﻿ - ﻿$y^{2}$﻿

d)

4 - ﻿$x^{2}$﻿

e)

﻿$x^{2}$﻿ - 81

20) ﻿$(2 \ + \ x)(x \ - \ 2)$﻿

a)

﻿$x^{2}$﻿ - 4

b)

81 - ﻿$x^{2}$﻿

c)

﻿$x^{2}$﻿ - ﻿$y^{2}$﻿

d)

4 - ﻿$x^{2}$﻿

e)

﻿$x^{2}$﻿ - 81

21) ﻿$(x \ + \ y)(x \ - \ y)$﻿

a)

﻿$x^{2}$﻿ - 4

b)

81 - ﻿$x^{2}$﻿

c)

﻿$x^{2}$﻿ - ﻿$y^{2}$﻿

d)

4 - ﻿$x^{2}$﻿

e)

﻿$x^{2}$﻿ - 81

##### Problem 6

Four students solved the equation ﻿$x^{2} + 225 = 0$﻿. Their work is shown here. Only one student solved it correctly.

Student A:

﻿$\begin{array}{rcl} \\[-1em] x^2 + 225 &=& 0 \\[-1em] \\ \\[-1em] x^2 &=& -225 \\[-1em] \\ \\[-1em] x = 15 & \text{or} & x = -15 \end{array}$﻿

Student B:

﻿$\begin{array}{c} \begin{array}{rcl} \\[-1em] x^2 + 225 &=& 0 \\[-1em] \\ \\[-1em] x^2 &=& -225 \\[-1em] \\ \end{array} \\ \text{No Solutions} \end{array}$﻿

Student C:

﻿$\begin{array}{rcl} \\[-1em] x^2 + 225 &=& 0 \\[-1em] \\ \\[-1em] (x - 15)(x + 15) &=& 0 \\[-1em] \\ \\[-1em] x = 15 &\text{or}& x = -15 \\[-1em] \end{array}$﻿

Student D:

﻿$\begin{array}{rcl} \\[-1em] x^2 + 225 &=& 0 \\[-1em] \\ \\[-1em] x^2 &=& 225 \\[-1em] \\ \\[-1em] x = 15 &\text{or}& x = -15 \\[-1em] \end{array}$﻿

Determine which student solved the equation correctly. For each of the incorrect solutions, explain the mistake.

22) Which student solved the equation correctly?

23) If Student A was incorrect, explain Student A's mistake.

24) If Student B was incorrect, explain Student B's mistake.

25) If Student C was incorrect, explain Student C's mistake.

26) If Student D was incorrect, explain Student D's mistake.