Different Methods of Checking Solutions of Quadratic Equations

9.7.3 Different Methods of Checking Solutions of Quadratic Equations

9.7.3 • Different Methods of Checking Solutions of Quadratic Equations

Activity

1. The equation $h (t) = 2 + 30 t - 5 t^{2}$ 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. The pumpkin reached a maximum height of 47 meters. How many seconds after launch did that happen?

3. The pumpkin reached a maximum height of 47 meters 3 seconds after launch.

b. Show your reasoning.

Compare your answer:

$\begin{matrix} h (t) = - 5 t^{2} + 30 t + 2 \\ - 5 t^{2} + 30 t + 2 = 47 \\ - 5 t^{2} + 30 t - 45 = 0 \\ a = - 5, b = 30, c = - 45 \\ t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ t = \frac{- (30) \pm \sqrt{(30)^{2} - (4) (- 5) (- 45)}}{2 (- 5)} \\ t = \frac{- 30 \pm \sqrt{900 - 900}}{- 10} \\ t = \frac{- 30}{- 10} \\ t = 3 \end{matrix}$

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

Compare your answer:

Substitute 3 for $t$ in the original expression; evaluating it gives $2 + 30 (3) - 5 (3)^{2}$ or $2 + 90 - 45$ , which is 47.
Graphing $y = 2 + 30 t - 5 t^{2}$ and $y = 47$ shows an intersection is located at $(3, 47)$ .
Graphing $y = - 45 + 30 t - 5 t^{2}$ shows a zero at $(3, 0)$ .

2. The equation $r (p) = 80 p - p^{2}$ 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.

Compare your answer:

Partially agree. A ticket price of $15.50 will generate about $1000 in revenue, but a ticket price of $74.50 will generate much less (about $410).
Using the quadratic formula to solve $1000 = 80 p - p^{2}$ gives $p \approx 15.50$ and $p \approx 64.50$ as the solutions.
Substituting 15.50 into the expression $80 p - p^{2}$ gives 999.75, which is close to 1000. $(80 (15.50) - (15.50) 2 = 999.75)$ . Substituting 74.50 into the expression $80 p - p^{2}$ gives 409.75.
Graphing the equation $y = r (p)$ and $y = 1000$ shows two intersections at approximately $(15.50, 1000)$ and $(64.50, 1000)$ .

Are you ready for more?

Extending Your Thinking

Function $g$ is defined by the equation $g (t) = 2 + 30 t - 5 t^{2} - 47$ . Its graph opens downward.

Find the zeros of function $g$ without graphing.

Compare your answer:

Function $g$ has one zero: 3.

Show your reasoning.

Compare your answer:

$2 + 30 t - 5 t^{2} - 47$ is equivalent to $- 5 t^{2} + 30 t - 45$ . Solving $- 5 t^{2} + 30 t - 45 = 0$ using the quadratic formula gives $t = 3$ .
$- 5 t^{2} + 30 t - 45$ can be rewritten as $- 5 (t^{2} - 6 t + 9)$ and then as $- 5 (t - 3) (t - 3)$ . Applying the zero product property to solve $- 5 (t - 3) (t - 3) = 0$ gives $t = 3$ .

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

Compare your answer:

Function $g$ only has one zero, which means its graph has only one horizontal intercept, occurring when $t = 3$ . Because the point $(3, 0)$ is the only intercept, it must also be the vertex of the graph, otherwise there would have been either two horizontal intercepts, or none.

Recall that $g (t) = 2 + 30 t - 5 t^{2} - 47$ and $h (t) = 2 + 30 t - 5 t^{2}$ . Examine these functions $g$ and $h$ (which defined the height of the pumpkin). Explain how the maximum of function $h$ , once we know it, can tell us the maximum of $g$ .

Compare your answer:

The value of the expression for $g$ is 47 less than that of $h$ , so if we know the maximum of $h$ , we know that the maximum of $g$ is 47 less than that. We saw that the maximum of $h$ is 47, so the maximum of $g$ is 47 less than 47, which is 0.

Video: Explaining Your Solutions to Quadratic Situations

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

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Self Check

The equation

g(t) = 2 + 23.7t - 4.9t^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?

$t=\frac{-23.7 \pm \sqrt{600.89}}{4}$
$t=\frac{-23.7 \pm \sqrt{600.89}}{-4.9}$
$t=\frac{-23.7 \pm \sqrt{600.89}}{-9.8}$
$t=\frac{23.7 \pm \sqrt{600.89}}{-9.8}$

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 - $(x + 3)^{2} + 9 = 0$

Step 2 - $(x + 3)^{2} = - 9$

Step 3 - $x + 3 = \pm \sqrt{- 9}$

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 $49 + x^{2} = 0$ 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

Answering Questions About Situations With the Quadratic Formula

Function $h$ models the height of an object, in meters, $t$ seconds after it is launched into the air. It is defined by $h (t) = - 5 t^{2} + 25 t$ . How much time will it take the object to reach 15 meters?

Here is how to solve this word problem using the quadratic formula:

To know how much time it would take the object to reach 15 meters, we could solve the equation $15 = - 5 t^{2} + 25 t$ .

Rewriting it in standard form gives $- 5 t^{2} + 25 t - 15 = 0$ . The expression on the left side of the equation cannot be written in factored form, however.
Completing the square isn't convenient because the coefficient of the squared term is not a perfect square and the coefficient for the linear term is an odd number.
Let's use the quadratic formula, using $a = - 5$ , $b = 25$ , and $c = - 15$ !

Step 1 - $t = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

Step 2 - $t = \frac{- 25 \pm \sqrt{25^{2} - 4 (- 5) (- 15)}}{2 (- 5)}$

Step 3 - $t = \frac{- 25 \pm \sqrt{325}}{- 10}$

The expression $\frac{- 25 \pm \sqrt{325}}{- 10}$ represents the two exact solutions of the equation.

We can get approximate solutions by using a calculator, or by reasoning the $\sqrt{325} \approx 18$ .

The solutions tell us that there are two times after the launch when the object is at a height of 15 meters: at about 0.7 seconds (as the object is going up) and 4.3 seconds (as it comes back down).