Quadratics

3.2 (OPTIONAL) An Aside on Imaginary Solutions

Be sure to have looked at PART 2 of this QUADRATICS course so that this section makes sense to you!

 

As we have seen, it is possible for a quadratic equation to have no solutions. For example:

 

EXAMPLE: Solve \(9+z^{2}=4z\).

 

Answer: Let’s bring all terms containing the variable to one side.

\(z^{2}-4z+9=0\).

 The box method shows we should subtract \(5\) from both sides. (Verify this!)

\(z^{2}-4z+4=-5\)

\((z-2)^{2}=-5\)

 And this has no solution in the realm of real numbers: No number squared will equal negative five.

 

However, mathematicians, after centuries of confused thought, have finally settled that it can be very useful in advanced work to imagine there is a number with the property of its square being negative. They call that number \(i\) and declare \(i^{2}=-1\).

So \(\sqrt{-1} = i\) ( or should that be \(-i\)?  Hmm!) and \(\sqrt{-5}=\sqrt{-1 \times 5} =i\sqrt{5}\), and so on.

 

From \((z-2)^{2}=-5\)  we can now keep going and write:

\(z-2=i\sqrt{5}\)    or    \(-i\sqrt{5}\)

\(z=2+i\sqrt{5}\)    or    \(2-i\sqrt{5}\)  □

 

There is good reason why mathematicians want to allow imaginary numbers in their work, but those reasons don’t arise until one studies advanced trigonometry, calculus, and advanced algebra systems. In high-school algebra course there is no reason to ever want to consider imaginary solutions to equations.

 

HOWEVER, many curricula feel that is “good for students” to see advanced ideas in mathematics before they will ever need them. Why? I don’t know. In fact, I personally feel it is bad to bring in ideas early simply because you “might need to know it later on.”

 

One more example to make these algebra curriculum writers happy:

 

EXAMPLE: Soleve \(2x^{2}+2x+1=0\).

 

Answer: Multiply through by 2:

\(4x^{2}+4x+2=0\).

 Use the box method to see that the magic number we seek is “\(1\)”:

\(4x^{2}+4x+1=-1\)

\((2x+1)^{2}=-1\)

 This has no solution within the reals. Within the system of imaginary numbers we can continue on:

 \(2x+1=i\)  or  \(-i\)

yielding

\(x=\frac{-1+i}{2}\)   or  \(\frac {-1-i} {2}\).   □

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