Complex Numbers in Quadratic EquationsDate: 11/09/1999 at 10:08:17 From: Takeyuki Oya Subject: Imaginary numbers in solving quadratic equations Hello, I just want to know about the use of imaginary numbers in solving quadratic equations, particularly when the discriminant is negative. Also if possible, could you give me a description of how solutions of this type of quadratic equation may be represented graphically? Thank you. Date: 11/09/1999 at 17:06:24 From: Doctor Rick Subject: Re: Imaginary numbers in solving quadratic equations Hello, welcome to Ask Dr. Math. You've asked some good questions. >I just want to know about the use of imaginary numbers in solving >quadratic equations, particularly when the discriminant is negative. Let's take an example: x^2 - 4x + 7 = 0 Using the quadratic equation, we get x = (1/2)(4 +- sqrt(16-28)) = (1/2)(4 +- sqrt(-12)) What is the square root of -12? We can factor -12 = -1 * 3 * 4. Then sqrt(-1*3*4) = sqrt(-1)*sqrt(3)*sqrt(4) = i * sqrt(3) * 2 using the definition of the imaginary number i = sqrt(-1). Now we can continue simplifying x: x = (1/2)(4 +- 2i*sqrt(3)) = 2 +- i*sqrt(3) This is our final answer: the roots of the quadratic are x = 2 + i*sqrt(3) and x = 2 - i*sqrt(3) Whenever the discriminant is negative, its square root will be imaginary. The real part of the two solutions is always the same. Can you see why this is true? We can write out the general solution for the case in which the discriminant is negative: ax^2 + bx + c = 0 x = (-b +- sqrt(b^2 - 4ac))/(2a) = (-b +- i*sqrt(4ac - b^2))/(2a) The real part is -b/(2a), and the imaginary part is +-sqrt(4ac-b^2)/(2a). >Also if possible, could you give me a description of how solutions of >this type of quadratic equation may be represented graphically? In order to see the complex roots on a graph, we need to see complex values of x. Therefore we'll need to replace the real x-axis with a complex plane. So, draw a plane with two axes: Re(x), the real part of x, and Im(x), the imaginary part of x. Then "draw" the y-axis perpendicular to this plane. (You'll have to either imagine this axis coming out of the paper, or draw it in perspective.) Any point in the x-plane corresponds to a complex value of x; we can evaluate y = ax^2 + bx + c and plot this point (Re(x),Im(x),y). We will end up with a surface rather than a curve: for any point in the x-plane, the surface will be at height y(Re(x),Im(x)) above the plane. Wait a moment - did you notice something wrong there? If x can be any complex number, then y(x) may be complex itself. I was talking as if the height y(x) were a real number. What can we do about this? One way to picture it is to look at the real and imaginary parts of y(x) separately. In other words, you can draw two surfaces, one for the real part of y and the other for the imaginary part of y. We're mainly interested in the roots of the quadratic - where the "surface" y(x) crosses the x-plane. This will only happen when BOTH the real and imaginary parts of y are zero. I won't try to describe all the work that is needed; this makes an interesting project. I suggest that you write the complex number x as x_r + i*x_i where x_r (the real part of x) and x_i (the imaginary part of x) are both real. Substitute this in the quadratic: y = (x_r + i*x_i)^2 + b(x_r + i*x_i) + c Separate y into real and imaginary parts. Think about what each surface will look like. In particular, consider: for which values of x_r and x_i will the imaginary part be zero? Then consider how the real part of y behaves for these values of x_r and x_i. You will discover the real parabola that you expect, plus something interesting! I've probably given you more than you wanted, but this really is an interesting question to pursue. I hope you can try it. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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