Completing the Square (Part 2)

ID: juriz-fijin
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Created by Illustrative MathematicsIllustrative Mathematics, CC BY 4.0
Subject: Algebra, Algebra 2
Grade: 8-9
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26 questions

Completing the Square (Part 2)

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Due:
Student Name:
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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 x2+3xx^{2} + 3x in order to write it as a perfect square.

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2) Write an equivalent expression in factored form.

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3) Add a number to x2+0.6xx^{2} + 0.6x in order to write it as a perfect square.

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4) Write an equivalent expression in factored form.

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5) Add a number to x211xx^{2} - 11x in order to write it as a perfect square.

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6) Write an equivalent expression in factored form.

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7) Add a number to x252xx^{2} - \frac{5}{2}x in order to write it as a perfect square.

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8) Write an equivalent expression in factored form.

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9) Add a number to x2+xx^{2} + x in order to write it as a perfect square.

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10) Write an equivalent expression in factored form.

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Problem 2

11) Noah is solving the equation x2+8x+15=3x^2 + 8x + 15 = 3. He begins by rewriting the expression on the left in factored form and writes (x+3)(x+5)=3(x + 3)(x + 5) = 3. He does not know what to do next.

Noah knows that the solutions are x=2x = -2 and x=6x = -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) x2+20x+50=14x^{2} + 20x + 50 = 14. The solutions are x=18x = -18 and x=2x = -2.

13) x2+1.6x=0.36x^{2} + 1.6x = 0.36. The solutions are x=1.8x = -1.8 and x=0.2x = 0.2.

14) x25x=114x^{2} - 5x = \frac{11}{4}. The solutions are x=112x = \frac{11}{2} and x=12x = \frac{-1}{2}.

Problem 4

Solve each equation.

15) x20.5x=0.5x^{2} - 0.5x = 0.5

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16) x2+0.8x=0.09x^{2} + 0.8x = 0.09

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17) x2+133x=5636x^{2} + \frac{13}{3}x = \frac{56}{36}

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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)(2 \ + \ x)(2 \ - \ x)

a)

x2x^{2} - 4

b)

81 - x2x^{2}

c)

x2x^{2} - y2y^{2}

d)

4 - x2x^{2}

e)

x2x^{2} - 81

19) (x + 9)(x  9)(x \ + \ 9)(x \ - \ 9)

a)

x2x^{2} - 4

b)

81 - x2x^{2}

c)

x2x^{2} - y2y^{2}

d)

4 - x2x^{2}

e)

x2x^{2} - 81

20) (2 + x)(x  2)(2 \ + \ x)(x \ - \ 2)

a)

x2x^{2} - 4

b)

81 - x2x^{2}

c)

x2x^{2} - y2y^{2}

d)

4 - x2x^{2}

e)

x2x^{2} - 81

21) (x + y)(x  y)(x \ + \ y)(x \ - \ y)

a)

x2x^{2} - 4

b)

81 - x2x^{2}

c)

x2x^{2} - y2y^{2}

d)

4 - x2x^{2}

e)

x2x^{2} - 81

Problem 6

Four students solved the equation x2+225=0x^{2} + 225 = 0. Their work is shown here. Only one student solved it correctly.

Student A:

x2+225=0x2=225x=15orx=15\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:

x2+225=0x2=225No Solutions\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:

x2+225=0(x15)(x+15)=0x=15orx=15\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:

x2+225=0x2=225x=15orx=15\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.