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Algebra 1

9.7.3 Different Methods of Checking Solutions of Quadratic Equations

Algebra 19.7.3 Different Methods of Checking Solutions of Quadratic Equations

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Activity

1. The equation h(t)=2+30t5t2h(t)=2+30t5t2 represents the height, as a function of time, of a pumpkin that was catapulted up in the air. Height is measured in meters and time is measured in seconds.

a. Enter the number of seconds after launch.

The pumpkin reached a maximum height of 47 meters. How many seconds after launch did that happen?

b. Show your reasoning.

c. Suppose someone was unconvinced by your solution. Find another way (besides the steps you already took) to show your solution is correct.

2. The equation r(p)=80pp2r(p)=80pp2 models the revenue a band expects to collect as a function of the price of one concert ticket. Ticket prices and revenues are in dollars.

A band member says that a ticket price of either $15.50 or $74.50 would generate approximately $1000 in revenue. Do you agree? Show your reasoning.

Video: Explaining Your Solutions to Quadratic Situations

Watch the following video to learn more about proving quadratic solutions.

Are you ready for more?

Extending Your Thinking

1.

Function gg is defined by the equation g(t)=2+30t5t247g(t)=2+30t5t247. Its graph opens downward.

Find the zeros of function gg without graphing.

2.

Show your reasoning.

3.

Explain or show how the zeros you found can tell us the vertex of the graph of gg.

4.

Recall that g(t)=2+30t5t247g(t)=2+30t5t247 and h(t)=2+30t5t2h(t)=2+30t5t2. Examine these functions gg and hh (which defined the height of the pumpkin). Explain how the maximum of function hh, once we know it, can tell us the maximum of gg.

Self Check

The equation g ( t ) = 2 + 23.7 t 4.9 t 2 models the height, in meters, of a pumpkin t seconds after it has been launched from a catapult. Which of the following substitutions into the quadratic formula is correct to find the value of t where the pumpkin hit the ground?
  1. t = 23.7 ± 600.89 4
  2. t = 23.7 ± 600.89 4.9
  3. t = 23.7 ± 600.89 9.8
  4. t = 23.7 ± 600.89 9.8

Additional Resources

Answering Questions About Situations With the Quadratic Formula

Here is an example of someone solving a quadratic equation that has no solutions:

Step 1 - (x+3)2+9=0(x+3)2+9=0

Step 2 - (x+3)2=9(x+3)2=9

Step 3 - x+3=±9x+3=±9

Study the example. At what point did you realize the equation had no solutions?

  • For two numbers to add up to 0, they have to be opposites. Since the number 9 is positive, the other number must be negative, which a square cannot be.

Explain how you know the equation 49+x2=049+x2=0 has no solutions.

  • A square is always positive, so a square plus a positive number must also be positive and can never equal zero.

Try it

Try It: Answering Questions About Situations With the Quadratic Formula

Function hh models the height of an object, in meters, tt seconds after it is launched into the air. It is defined by h(t)=5t2+25th(t)=5t2+25t. How much time will it take the object to reach 15 meters?

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