After linear equations and exponential equations, the next most common kind of equations are quadratic equations, which we mentioned briefly in Section 2.3. Quadratic equations are made up of a power equation with power 2, added to a linear equation. In this section we’ll learn about how to solve quadratic equations using our friend the quadratic formula.
A good way to get things to stick in your brain is to try to remember them without looking them up. Even if you get them wrong at first, the effort of trying to recall them is helpful!
Huh? I thought the beanbag went up in the air. What’s happening here? Oh, I know. The beanbag must be falling down by then. As we fill in the table with intermediate values we see how Claude’s beanbag went up in the air and then back down.
Notice by \(T=1.2\) seconds we got \(-2.04\) feet. We can’t have negative feet. The beanbag must hit the ground before 1.2 seconds. From the graph I’d say in just over 1.1 seconds.
Of course, we can refine our answer by successive approximations. The beanbag hits the ground when its height is 0 feet. Looks a little strange but we want \(H =0\text{.}\) We expect the answer is just a little bigger than 1.1, so we start our guess optimistically with 1.11.
In this chapter we’ve seen how to solve linear, power, and exponential equations. Let’s solve this equation too. By the way, our function is quadratic because
We want to solve for \(T\text{.}\) Notice that because \(T\) occurs twice in the equation, nothing we have seen to do to each side of the equation can knock it down to just one \(T\text{.}\) That means none of our methods so far work. Luckily there’s a way to solve any quadratic equation using the aptly-named Quadratic Formula.
Oh my! First thing to understand in this complicated formula is that because of the “\(\pm\)”, which we pronounce “plus or minus”, we actually get two possible answers:
\begin{equation*}
T = \frac{-b}{2a}+ \frac{\sqrt{b^2-4ac}}{2a}
\quad \text{and} \quad
T = \frac{-b}{2a} - \frac{\sqrt{b^2-4ac}}{2a}\text{.}
\end{equation*}
Sometimes only one answer makes sense in the story, other times they both might. Stay tuned.
As usual, we needed parentheses around the denominator (bottom) of our fraction to override the normal order of operations. As a reminder, (-) refers to negation. Remember, this does not mean you type in parentheses, just hit the one key that is labeled already.
Check out the parentheses now. Three sets here. First, around the quantity we’re taking the square root of. Maybe your calculator included the open parentheses along with the square root, but either way we need them. Second, around the number (15) that we are squaring. That didn’t matter here but if \(b\) were negative it would have. Last, we added parentheses around the bottom of the fraction, as always.
We had guessed around 1.106 seconds, so that is definitely the right answer: Claude’s beanbag will hit the ground after 1.10689 seconds. Yeah, too precise. But you get the idea.
Wait a minute! Any good juggler isn’t about to let the beanbag fall on the ground. He’s going to catch it again, perhaps at about \(3\tfrac12\) feet above ground. That means we’re looking for \(H=3.5\text{.}\) Using our equation \(H=3+15T-16T^2\text{,}\) we get
The Quadratic Formula only works if the equation has \(=0\text{,}\) but we have \(=3.5\text{.}\) It might seem that we’re out of luck, but it’s an easy fix. Just subtract 3.5 from each side.
Both answers seem to make sense. Let’s look at the graph to confirm that they’re reasonable. We first find \(H=3.5\) while the beanbag is going up in the air, just before the unlabeled gridline for 0.05 (midway between 0 and 0.1) so an answer of \(T\approx 0.03\) makes sense. Then, on the way back down, the beanbag is 3.5 feet up at what looks like 0.9 seconds, so our answer of \(T \approx 0.9\) makes sense too. Since Claude catches the beanbag on the way down, we want that second answer, after 0.90289 seconds (which is just before 1.106 seconds when it hits the ground, by the way).
High jumpers actually jump into a pit, which is like a big soft beanbag that’s about 2 feet thick. Use the graph above to estimate when the high jumper would hit the pit.
One interesting note. What happens in the story at the point where the beanbag stops going up in the air and starts falling down? That must be when the beanbag is at its highest point. What is the speed at that highest point? Well, I guess 0. For a split second it’s almost frozen in midair, neither rising nor falling. (If we were able to compute the rate of change for a really, really small interval of time then we would find the rate of change \(\approx 0\text{.}\))
In general, the graph of \(H = aT^2+bT+c\) is a parabola. The two solutions from the Quadratic Formula are both places where \(H=0\) (we give them the fancy name roots) and, so, the graph crosses the \(T\)-axis there. Might not make sense in the real problem, but the equation and formula don’t know that. (Okay, equations and formulas don’t actually “know” anything. But you get my point.) Turns out the graph is symmetric about the highest point, so that must be midway in between the roots which is exactly where
\begin{equation*}
T =\frac{-b}{2a}
\end{equation*}
Because \(a\) is negative, the answer we got by adding is to the left and the answer we got by subtracting is to the right.
Our graph was a \(\cap\) shaped parabola and so we found a maximum value. The graph of a quadratic function might be \(\cup\) shaped instead. In that case evaluating at \(T=\frac{-b}{2a}\) would give the minimum value.
The high jump pit is 2 feet off the ground. When does the high-jumper land in the pit? Use the Quadratic Formula to find the answer and the graph to check.
How high a bar can the high-jumper clear? Find the maximum height of that point above ground by evaluating at \(\displaystyle T=\frac{-b}{2a}\text{.}\) Use the graph to check.
The art museum opened in 1920. After an initial rush to see the great holdings, attendance dropped for awhile. But then attendance began to rise again and has risen since. The number of annual visits \(N\) is approximated by the equation
According to this equation, in what year was the number of annual visits the smallest? For that year, what were the number of visits? Use \(\displaystyle T=\frac{-b}{2a}\text{.}\)
How many tanks of milk would they need to sell to keep profits over $400,000? Set up and solve a quadratic equation to find the answer. Then check that it agrees with your graph. Your answer should be in the form of an inequality.
What was once an abandoned lot down the block is now a thriving 10′\(\times\)25′ vegetable and berry garden for the neighborhood. One neighbor volunteered to donate gravel to make a path around the garden. The path will be 3 inches deep and the same width all around.
If the neighbor donates 60 cubic feet of gravel, how wide a path can they build? Set up and solve a quadratic equation to find the answer in feet, accurate to two decimal places. Then convert your answer into inches.
Gravel is measured by the yard, which is short for cubic yard. There are 27 cubic feet in 1 yard of gravel. If the neighbor donates three yards of gravel, how wide a path can they build? Set up and solve a quadratic equation to find the answer in feet, accurate to two decimal places. Then convert your answer into inches.
Claude is an excellent juggler. Remember that the height \(H\) feet of Claude’s beanbag \(T\) seconds after he throws it in the air is described by the equation \(H = 3+15T-16T^2\text{.}\) Answer each of the following question by the suggested method and then look back at the graph from earlier to make sure your answers make sense.
When is the beanbag 8 feet above ground? Try to use the Quadratic Formula to find the answer. What happens? Explain why it makes sense in the story that you can’t solve this quadratic equation.
Claude decided that the beanbag was too high in the air, so he modified his throw slightly. Now the height is given by \(H = 3+14T-16T^2\text{.}\) What is the maximum height the beanbag will reach now? Hint: what number can you evaluate at?
The stopping distance for Jeff’s Cadillac Escalade is given by
\begin{equation*}
D=0.04S^2+1.47S
\end{equation*}
Aside
where \(S\) is the speed of the car (in miles per hour) and \(D\) is the stopping distance (in feet). Jeff took 183 feet to stop. How fast was he going?
Mrs. Weber’s cooking class came up with the equation
\begin{equation*}
M = 1.2F^2+4F+7
\end{equation*}
Aside
to approximate the grilling time of a piece of fish depending on its thickness. Here \(M\) is the number of minutes to grill the fish and \(F\) is the thickness of the fish in inches.
An analyst had suggested that they close down shop earlier, once sales were below $50 billion. In what year did sales fall that low? Show how to solve using the Quadratic Formula.
