8.5 - Influence of Zero [FOM11]

Fill in the Blank

Figure out the value of x that makes the equation true:

3x = 27

Fill in the Blank

Figure out the value of x that makes the equation true:

x³ = 27

Fill in the Blank

Figure out the value of x that makes the equation true:

3x = 24

How do we know the value of x that makes a solution?

How do you decide which operations to use in order to solve for x?

What are equivalent equations?

Categorize

Options (6)

$$7x-105+3=108$$  

$$7\left(x-15\right)-5=0$$  

$$7x-102\ -\ 8\ =\ 0$$  

$$7x-15+3=8$$  

$$7x-105+3=0$$  

$$7x-102-5=0$$  

Which equation(s) is/are equivalent to:

7(x - 15) + 3 = 8

Equivalent Equations

Non-Equivalent Equations

Fill in the Blank

Find the value of the equation when x = 6

$$f\left(x\right)=\left(x+6\right)\left(x-6\right)$$

Fill in the Blank

Find the value of the equation when x = 6

$$f\left(x\right)=\left(x+4\right)\left(x-6\right)$$

Fill in the Blank

Find the value of the equation when x = 6

$$f\left(x\right)=x^2-6x$$

Fill in the Blank

Find the value of the equation when x = 6

$$f\left(x\right)=2\left(\frac{2}{3}x+8\right)\left(x-6\right)$$

Fill in the Blank

Find the value of the equation when x = 6

$$f\left(x\right)=\frac{1}{2}x^2-3x$$

Open Ended

What do all of these responses have in common?

Fill in the Blank

Find another value of x so that f(x) = 0, as it was earlier when x = 6. (If there is no other value, write NONE)

$$f\left(x\right)=\left(x+6\right)\left(x-6\right)$$

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Fill in the Blank

Find another value of x so that f(x) = 0, as it was earlier when x = 6. (If there is no other value, write NONE)

$$f\left(x\right)=\left(x+4\right)\left(x-6\right)$$

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Fill in the Blank

Find another value of x so that f(x) = 0, as it was earlier when x = 6. (If there is no other value, write NONE)

$$f\left(x\right)=x^2-6x$$

Fill in the Blank

Find another value of x so that f(x) = 0, as it was earlier when x = 6. (If there is no other value, write NONE)

$$f\left(x\right)=2\left(\frac{2}{3}x-8\right)\left(x-6\right)$$

Fill in the Blank

Find another value of x so that f(x) = 0, as it was earlier when x = 6. (If there is no other value, write NONE)

$$f\left(x\right)=\frac{1}{2}x^2-3x$$

How can you find the value(s) of x that make the equation true for equation(s) in factored form like the one below?

(x - 10)(x + 3) = 0

What value(s) of x has to be a solution for equations without a constant like the one below?

x² - 7x = 0

Why doesn't the approach you took on #1 work for the equation below?

(x - 10)(x + 3) = 1

For each equation, state whether or not the solution is valid.

Multiple Choice

2(x - 1) + 4 = 3x - 2

2x - 2 + 4 = 3x - 2

2x = 3x

-x = 0

x = 0

• Correct!

• Incorrect!

Multiple Choice

3(x - 1) = 5x + 6

3x - 1 = 5x + 6

-1 = 2x + 6

-7 = 2x

-3.5 = x

• Correct!

• Incorrect!

Multiple Choice

(x - 2)(x + 3) = x + 10

x² + x - 6 = x + 10

x² - 6 = 10

x² = 16

x = 4

• Correct!

• Incorrect!

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Complete the 8.5 Practice Problems independently or with a partner for a prepare grade (due Wednesday, 5/8)