# Lesson 5

How Many Solutions?

• Let’s use graphs to investigate quadratic equations that have two solutions, one solution, or no 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. $$6x(x-8) = 29$$

### 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.

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

$$g(x) = x(x+3)$$

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

2. Determine how many solutions each $$f(x)=0, g(x)=0$$, and $$h(x)=0$$ has. Explain how you know.

### 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.

### 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.

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

(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 $$x=6$$

(From Unit 7, Lesson 4.)

### Problem 6

1. Solve the equation $$25=4z^2$$.
2. Show that your solution or solutions are correct.
(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.
(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. $$(x+3)(x+7)=0$$