Lesson 5

How Many Solutions?

Problem 1

Rewrite each equation so that the expression on one side could be graphed and the \(x\)-intercepts of the graph would show the solutions to the equation.

  1. \(3x^2 = 81\)
  2. \((x-1)(x+1) -9 = 5x\)
  3. \(x^2 -9x + 10 = 32\)
  4. \(6x(x-8) = 29\)

Solution

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Problem 2

  1. Here are equations that define quadratic functions \(f, g\), and \(h\). Sketch a graph, by hand or using technology, that represents each equation.

    \(f(x)=x^2+4\)

    Blank coordinate grid, origin O.

    \(g(x) = x(x+3)\)

    Blank coordinate grid, origin O.

    \(h(x)=(x-1)^2\)

    Blank coordinate grid, origin O.
  2. Determine how many solutions each \(f(x)=0, g(x)=0\), and \(h(x)=0\) has. Explain how you know.

Solution

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Problem 3

Mai is solving the equation \((x-5)^2=0\). She writes that the solutions are \(x=5\) and \(x=\text- 5\). Han looks at her work and disagrees. He says that only \(x=5\) is a solution. Who do you agree with? Explain your reasoning.

Solution

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Problem 4

The graph shows the number of square meters, \(A\), covered by algae in a lake \(w\) weeks after it was first measured.

In a second lake, the number of square meters, \(B\), covered by algae is defined by the equation \(B = 975 \boldcdot \left(\frac{2}{5}\right)^w\), where \(w\) is the number of weeks since it was first measured.

Graph of function.

For which algae population is the area decreasing more rapidly? Explain how you know.

Solution

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(From Unit 5, Lesson 6.)

Problem 5

If the equation \((x-4)(x+6)=0\) is true, which is also true according to the zero product property?

A:

only \(x - 4 = 0\)

B:

only \(x + 6 = 0\)

C:

\(x - 4 = 0\) or \(x + 6 = 0\)

D:

\(x=\text-4\) or \(x=6\)

Solution

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(From Unit 7, Lesson 4.)

Problem 6

  1. Solve the equation \(25=4z^2\).
  2. Show that your solution or solutions are correct.

Solution

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(From Unit 7, Lesson 3.)

Problem 7

To solve the quadratic equation \(3(x-4)^2 = 27\), Andre and Clare wrote the following:

Andre

\(\displaystyle \begin {align} 3(x-4)^2 &= 27 \\ (x-4)^2 &= 9 \\ x^2 - 4^2 &= 9 \\ x^2 - 16 &= 9 \\ x^2 &= 25 \\ x = 5 \quad &\text{ or }\quad x = \text- 5\\ \end {align}\)

Clare

\(\displaystyle \begin{align} 3(x-4)^2 &= 27\\ (x-4)^2 &= 9\\ x-4 &= 3\\ x &= 7\\ \end{align}\)

  1. Identify the mistake each student made.
  2. Solve the equation and show your reasoning.

Solution

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(From Unit 7, Lesson 3.)

Problem 8

Decide if each equation has 0, 1, or 2 solutions and explain how you know.

  1. \(x^2 -144=0\)
  2. \(x^2 +144=0\)
  3. \(x(x-5)=0\)
  4. \((x-8)^2=0\)
  5. \((x+3)(x+7)=0\)

Solution

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