Carmen felt brushed off
Carmen had asked Ms. Clementine, the teacher, a question about quadratic equations. She felt brushed off because she had been told that the teacher would deal with it in another class. When would she deal with it? All the lessons on the quadratic function seemed to be over. Carmen was curious but did not like to study things on her own. She ran into Ms. Clementine in the hallway and again asked her about her question. The teacher told her that she would deal with it today, and then they both came to the class.
Ms. Clementine: Today, I want to deal with a question which Carmen had asked in a previous class when we were talking about quadratic equations. You remember, the quadratic equations can be solved by many different methods but Sara had told us that we can solve y = ax2 + bx + c = 0 by using the formula x = (-b ± (√(b2-4ac)))/2a .
5 minutes to solve an equation
I will write an equation on the board and give you 5 minutes to solve it for x. Then she wrote:
y = x2 + 3x + 3 = 0.
After 5 minutes, Ms. Clementine: Who has the answer ?
Sara: I have the answer but it’s kind of funny. I have x = -3/2 ± (√-3)/2.
Ms. Clementine asked if anyone else had the same answer and many students said yes.
Ms. Clementine: Kathy, you showed us how to solve a quadratic equation by graphing. Can you do that for y = x2 + 3x + 3 = 0 ? Some of the students can help you because they have calculators.
With the help of other students, Kathy made several graphs like this one but in none of them the curve crossed the X-axis (Fig. 8.1). That meant that there was no value of x at which y = x2 + 3x + 3 = 0.
Ms. Clementine: Kathy, so would it be right to say from your graph that there is no real solution to the equation y = x2 + 3x + 3 = 0. Sara, how do you explain your funny solution then ?
Imaginary numbers in electromagnetism
Sara: We ran into an idea in our physics class while studying alternating current or as we say AC in short. We normally say that the current is the ratio of voltage to resistance. However, anytime a current goes through a wire, it creates a magnetic field around it. For an alternating current, the direction of the magnetic field keeps changing. This field is in a plane perpendicular to the plane of the current, and the change in the direction of the field can slow the current down. Therefore, we talked of impedance which is the resistance plus the effects of the magnetic field. It was written as Z = R +jX and here j = √-1. The R is considered to be real and jX to be imaginary.
Introducing iota and its properties
Ms. Clementine: Very good Sara, here Z would be called complex number because it consists of two parts: R which is real and jX which is imaginary. I am sure your physics teacher will tell you much more about the reasoning behind it. Also, here in the Math class, instead of j we are going to use iota (i) which is √-1 .
Sara raised her hand and said: Ms. Clemetine, instead of x = -3/2 ± (√-3)/2, I should have said that x = -3/2 ± i√3/2.
Ms. Clementine: That’s what you could have said. Now I want to see how many of you understand the concept of iota. I will write some equations on the board. See if they make sense to you.
i2= -1, 1/i= –i, (a + ib) + (c + id) = a + c+ i(b + d), (a + ib)(a + ib) = a2 – b2 + 2iab,
(a +ib)(a –ib) = a2 + b2, 1/(i – 1) = (i +1)/((i – 1)(i + 1)) = (i+1)/(i2-1) = – (i+1)/2.
After a few more minutes she said: Joe, you showed us how to solve quadratic equations by factorization. Solve y = x2 + 3x + 3 = 0 by this method.
Joe looked puzzled but went to the board anyway. Finally with the help of the class, he came up with
y = x2 + 3x + 3 = (x + 3/2+ i√3/2) (x + 3/2 – i√3/2) = 0
Therefore, x = -3/2 + i√3/2 or x = -3/2 – i√3/2.
Ms. Clementine: Jun, you were so excited about finding the vertex of the Kwong’s quadratic equation. Can you come and show us how to do this for y = x2 + 3x + 3?
Jun: For y = ax2 + b + c, at the vertex 2ax + b = 0.
Here a = 1 and b = 3, therefore, the vertex is when 2 x + 3 = 0 or x = -3/2. That is when the value of y will be either maximum or minimum but Kathy’s graph showed that this point is the minimum. Then y will never really be zero in her graph.
Ms. Clementine: Carmen, I hope your question is answered by now.
Carmen: Yes, Ms. Clementine. Thank you very much. It was very interesting.
Challenge
Cube root of 1 is denoted by the Greek letter omega (ω). That means ω3 = 1, solve for ω.
Solution: ω3 = 1 can also be written as (ω3 – 1) = 0 or (ω – 1)( ω2 + ω + 1) = 0.
If (ω – 1) = 0, ω =1.
When ω2 + ω + 1 = 0, the quadratic formula gives ω = -1/2 ± (i√3)/2. This gives us the other two roots. Therefore, ω can be 1, (-1/2 + (i√3)/2) or -(1/2 – (i√3)/2).
Math Stories For You 