# Introduction to quadratic equations

Ms. Clementine: Kwong, now that you have brought out the concept of a quadratic relationship in your science fair project, maybe the whole class should learn about it. I was going to teach it later but this seems to be a good time. Because, the class has already participated in this survey, may be everyone can also talk about it. Sara, would you want to give the class a basic idea of a quadratic relationship?

Sara: I can try but only the very basic idea.

Ms Clementine: Please come to the front then.

Sara said, “Hi everybody. Like all of you, I still have to learn about this area. Do you remember when we learned about the linear relationship y = a + bx? For example, when you are walking, your position depends on from where you started plus how fast you walk times how long you have walked. I think everybody in class remembers that.

It is not always that simple. Not all relationships are linear. Kwong showed you a curved graph. Let’s assume that students can make good use of phones to learn. Maybe by using the smart phones, they can get more resources for learning or maybe explain things to each other. Then you would expect a positive effect of cell phone use on academics. But then, somebody could say that the time you spend on the phone is being taken away from your study time. If so, the phone use will decrease your study time or the amount of attention you can pay to your studies. Excessive use of phone will have negative results like Ms. Clementine pointed out to the class. With a quadratic equation you would be able to explain both the effects together. It will have the general form of y = ax^{2} + bx + c. In Kwong’s example, it was y = -2.4x^{2} +9.6x + 76.8. I guess, Ms. Clementine will teach us more about it.

Ms. Clementine: Everyone, you know that a quadratic function is written as y = ax^{2} + bx + c. Typically, you will be asked to determine the values of x when

ax^{2} + bx + c = 0.

# Solving the quadratic equations

I can think of four different ways of solving the equation: graphing, factorisation, completing the squares and using the quadratic formula. Each method will be presented by a different group of students. The next class will be on the first two methods, and the class after that will be for the third and fourth methods. After this, one group will present the concept of vertex. The remaining students will come prepared to report on quadratic relationships in something of their experience or interest. I will post the list of students and their topics.

# ———————-Next Class———————-

Ms. Clementine: Let’s have the presentation from the first group. Kathy: We used Kwong’s equation for the academic performance vs smart phone usage as shown in this picture. Because, we wanted to solve the equation -2.4x^{2} +9.6x + 76.8 = 0, we plotted a continuous graph over more values and over a larger domain of x than Kwong had showed us (Fig. 7.3). We then looked for the x values where y =0. The graph showed that y=0 when x = -4 or x = +8. Using phone for minus 4 hours/day didn’t make any sense. Negative marks in the graphs didn’t make any sense to us either. Therefore, we concluded that it is possible to get a grade average of zero if you use the phone for 8 hours per day. When do you eat, sleep, play games and study if you are using the phone for 8 hours a day?

Ms. Clementine: Thank you for a clear presentation. Group 2 is next.

Joe: We also solved the equation for Kwong’s quadratic function but by factorisation. The quadratic equation was -2.4x^{2} + 9.6x + 76.8 = 0. At first, we were puzzled about how to factorise this equation with all the decimals but then someone helped us. Thanks Sara. Sara told us that the problem becomes very easy if we divide both sides of the equation by -2.4. By doing so, the equation became x^{2} – 4x -32 = 0. Now, 32 = 4 × 8. Also 8 – 4 = 4, and 4 is the middle term on the left. Therefore, the factorised equation becomes (x-8) (x+4) = 0. That means either

x – 8 = 0 which gives x = 8 or x + 4 = 0 which gives x = -4. So we got the same answer as the first group but we didn’t think much of the meaning of the negative values of x.

Ms. Clementine: Very good. You have the same answer for the same equation. I will give you a sheet with 6 different equations. Everyone has to do them by the two methods presented by Kathy and Joe, and hand in the answers in the next class.

# ——————-Next Class——————-

Groups 3 and 4 presented in the next class.

Jonathan: We read in the book that, in completing the square, the idea is to re-arrange the quadratic equation terms so that you have only one term with x in it and the other term is a constant. Like the other group, we also divided Kwong’s equation by -2.4. That made it much easier. Then we had x^{2} – 4x -32 = 0. So we started thinking about working with x^{2} – 4x to make a square term.

We said that (x – 2)^{2} = x^{2} – 4x + 4. Therefore, x^{2} – 4x = (x -2)^{2 }– 4.

Therefore, we rewrote x^{2} – 4x -32 = 0 as (x – 2)^{2 }= 32 + 4 = 36.

This gave us x -2 = ±√36 or x – 2 = ± 6. That gave us x = -4 or +8. The same answer as the other groups.

Ms. Clementine: Very good. You got the same answer as the others. Now, group 4 will present.

Sara: We were supposed to derive the generalised formula for solving quadratic equations. Our group used the same method as group 3 – completing the square.

The general equation is ax^{2} + bx + c = 0. Like the other groups, we divided both sides by a, to get: x^{2} /a + xb/a + c/a = 0 or x^{2} /a + xb/a = – c/a.

Then we completed the square by adding (b/2a)^{2} to both sides.

This gave us x^{2} /a + xb/a + (b/2a)^{2} = – c/a + (b/2a)^{2 }which is the same as:

(x + b/2a)^{2 }= – c/a+(b/2a)^{2}.

By taking square root of both the sides we get:

x + b/2a = ±√(- c/a+(b/2a))^{2} = ±(√(b2-4ac))/2a

That gave us x = (-b ± (√(b2-4ac)))/2a

In this method, you can simply write the values of a, b and c and get the values of x when y =0.

Kwong’s equation is -2.4x^{2} + 9.6x + 76.8=0. We used: a = -2.4, b = 9.6 and c = 76.8 and got the answer x = 2 ± 6. Surprise, surprise. This is the same answer as everyone else’s.

Ms. Clementine: Very good. Sara, you made this derivation look very simple.

Carmen raised hands: What happens when b^{2 }– 4ac is negative?

Ms. Clementine: We will take up this issue in another class. Please wait until then.

Okay class, remember the sheet I gave you with 6 different equations in the last class. Everyone has to solve them by the methods shown by Jonathan and Sara, and hand in the answers next class.

# Vertex

Ms. Clementine: Group 5 presentation will start now.

Jun: I am Jun and I will speak on behalf of group 5. You all saw Kwong’s exciting graph that fitted the equation y = -2.4x^{2} + 9.6x + 76.8.

Everyone solved for x when y = -2.4x^{2} + 9.6x + 76.8 = 0.

I know that I don’t want to get a zero grade point average which is y here. I want to get the best mark I can. So enough of solving for y = 0 (she said it in a defying manner).

Group 1 gave us a graph in which the y value increases first and then it starts to decrease. In this graph, there is a value of x at which there is no increase or decrease in y. This is called the vertex.

In the graph, this peak occurs at x = 2. Now if x = 2, the value of y becomes -2.4 × 2^{2} + 9.6 × 2 + 76.8 which is 86.4. So we say the vertex is at x = 2 and y = 86.4 or at 2, y = 86.4. There is another way to do this. For y = ax^{2} + bx + c, the slope of the graph is 2ax + b. At the vertex, the slope which is 2ax + b = 0. This means the vertex is at x = -b/2a. We did the question this way and got the value of x to be -9.6/(-2.4×2) which is 2. From this we also get the value of y to be 86.4. So Kwong’s graph shows that for the best results, use the phone for 2 hours/day and improve your grade point average by almost 10%. Way to go Kwong.

Ms. Clementine: Thank you for the upbeat presentation Jun. Do you remember talking about vertices in geometry? A vertex is a point where two lines meet. Same way, for a parabola the vertex is a point where the rise and the falls meet. It could be when either a maximum or a minimum value will occur. This time you had a parabola in which you had a maxima. You can also get a parabola that decreases initially and then it starts to increase. In that case, the vertex for that parabola will be the minima for the curve. So the concept of vertex can be used to get maxima or the minima.

# Experience and interest of students

Okay, now we will have examples of quadratic relationships from the experience and interests of people. Please, one by one go in the same order in which I assigned this topic to you.

Ashley: I am the captain of the school’s cheerleading team. There are two situations quadratics will fit in. The first is somewhat like Kwong’s equation. Judges give higher scores if you start with an easy cheer, make the next one more difficult and keep doing this with for as many cheers as you can. Increasing the cheers gets you additional scores but every additional cheer decreases the overall performance of the team for all the cheers. This may be because the team members have difficulty learning too many cheers or because there is a time constraint. So this would give a quadratic relationship between the scores and the number of cheers. My friend Kim will tell you the other one.

Kim: We do a basket throw cheer. We throw a small girl into the air and then we all catch her. If we throw only a little bit, every one boos. If we throw very fast, we are scared that the head of that girl might hit the roof which is only 12 meters high. So, I have to figure out how fast to throw her so that her head reaches only a height of 11 meters to be safe. Sara told me that this would also be a problem of a quadratic relationship.

Dino: My mom loves vegetable gardening. In a 20 meter long row, she has 20 plants which give her 1000 tomatoes which is 50 tomatoes per plant. We got into an argument that she could get more tomatoes if she had more plants in the same 20 meter row. We asked a greenhouse and they told us that for each plant we add, the yield from every plant will decrease by one tomato. I think this is also a quadratic problem in which I have to find the vertex.

Tom: This is something that happened at my dad’s shop. A lady came in with a portrait picture with a height to width ratio of 4/3. She wanted the picture enlarged to 192 square inches, and wanted to know what the height and the width of the enlarged picture would be. My dad is really smart, and he told me that he figured it out using a quadratic equation.

Jorge: I play basketball but I am not very tall. I need to jump to dunk the ball so that my feet are 2.5 feet above the ground. The coach told me that his kinesiology teacher taught him the equation for jumping to the height to be 16t^{2 }-12t where t is time in seconds for which I am in the air. How long will I have to be in the air for being able to dunk?

Ms. Clementine: Great. So now you know how word problems related to quadratic equations come about. There are lots of them in your book. Your homework assignment is to do any six of them before the next class.

*Challenge*

Kim’s Problem: We do a basket throw cheer. We throw a small girl into the air and then we all catch her. If we throw only a little bit, every one boos. If we throw very fast, we are scared that the head of that girl might hit the roof which is only 12 meters high. So, I have to figure out how fast to throw her so that her head reaches only a height of 11 meters to be safe. Sara told me that this would also be a problem of a quadratic relationship.

Kim’s problem can be divided into two parts. The first part concerns the amount of time it will take for the cheerleader to reach the height of 11 m. The second part is about how fast to throw her. Both of these are easier to grasp by asking how long it will take her to reach the ground after dropping from the height of 11 m. So we will write the relationship

Distance = ut + at^{2}/2

Here the initial speed u = 0 because when the thrown girl reaches the top, her velocity must be zero because she is not moving up or down at that instance. Writing a = acceleration due to earth’s gravity which is 9.81 m/sec^{2}

Then 11 = 0 + 9.81 t^{2}/2 or t^{2}/2 = 11/9.81 or t^{2 }= 22/9.81 sec^{2 }= 2.2425 sec^{2 }or t = 1.4975 sec.

The velocity v with the downward motion will increase by the relationship

v = u + at or v = 0 + 9.81 x 1.4975 m/sec = 14.6908 m/sec.

To verify this as the correct answer, let’s write the relationship for the girl being thrown up with this initial velocity but note that the girl’s movement will now decelerate and hence will use the negative value of a

Distance = ut + at^{2}/2 or

Distance = 14.6908 m/sec x 1.4975 sec – (9.81 m/sec^{2 }x (1.4975 sec)^{2 })/2^{ }or

Which is 22 – 11 = 11 meters – the height Kim wants the girl to reach.

Dino’s Problem: My mom loves vegetable gardening. In a 20 meter long row, she has 20 plants which give her 1000 tomatoes which is 50 tomatoes per plant. We got into an argument that she could get more tomatoes if she had more plants in the same 20 meter row. We asked a greenhouse and they told us that the yield from every plant will decrease by one tomato for each plant we add. I think this is also a quadratic problem in which I have to find the vertex.

Let us assume that Dino’s mom adds x plants. Then she will have 20 + x plants. The yield of each plant was 50 tomatoes per plant but now it will decrease. Each added plant will decrease the yield by one tomato per plant or the yield will be 50 – x.

Thus yield (y) = (20 + x) x (50-x) = 1000 + 50x – 20x -x^{2}

y = – x^{2 }+ 30x+ 1000

Jun said in the class that vertex = 2ax + b = 0 for y = ax^{2} + bx + c

Using this relationship vertex = -2 x + 30 = 0 or x = 15.

Adding 15 plants will yield (20 + 15) x (50 – 15) = 1225 tomatoes.

At the vertex, the values can be either minimum or maximum. Obviously this is not a minimum value because the yield of 1225 tomatoes is greater than the current yield of 1000 tomatoes.

Tom’s Problem: This is something that happened at my dad’s shop. A lady came in with a portrait picture with a height to width ratio of 4/3. She wanted the picture enlarged to 192 square inches, and wanted to know what the height and the width of the enlarged picture would be. My dad is really smart, and he told me that he figured it out using a quadratic equation.

Let us say the width is x inches. Then the height will be 4x/3 and the area will be 4x^{2}/3. Therefore,

4x^{2}/3 = 192 or x^{2 }= 144 or x = ±12 inches of which only x = 12 inches makes sense.

So the width of the picture would be 12 inches and its height would be 12 times 4/3 or 16 inches. Verifying the area, 16 inches x 12 inches =192 square inches.

Jorge’s Problem: I play basketball but I am not very tall. I need to jump to dunk the ball so that my feet are 2.5 feet above the ground. The coach told me that his kinesiology teacher taught him the equation for jumping to the height to be 16t^{2 }-12t where t is time in seconds for which I am in the air. How long will I have to be in the air for being able to dunk?

This is similar to the cheerleader Kim’s problem except for the units. Everything is given here in feet. The relationship will become 16t^{2 }-12t = 2.5 or 16t^{2 }-12t – 2.5 =0.

For a quadratic equation ax^{2}+ bx + c = 0, x = (-b ± (√(b^{2 }-4ac))/2a

Therefore t = ((12 ± √ ( 12^{2 }– 4 x 16 x (-2.5)))/2 x 16 = 0.92 or minus 0.17 sec. Being in the air for negative time does not make sense. So Jorge will have to jump to be in the air for 0.92 sec for being able to dunk.