Equations with Imaginary Roots

Equations with imaginary roots

The student should now be able to solve and check equations which have imaginary roots.

Example. Solve x^2 - 4x + 13 = 0.

By the formula, x = \dfrac{4 \pm \sqrt{16 - 52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2} = \dfrac{4 \pm 6i}{2} = 2 \pm 3i.

Check: (2 + 3i)^2 - 4(2 + 3i) + 13 = 4 + 12i - 9 - 8 - 12i + 13 = 0. In the same way 2 - 3i checks.

An equation such as x^3 = 1 is solved by factoring: x^3 - 1 = 0, so (x - 1)(x^2 + x + 1) = 0, and the quadratic factor gives two imaginary roots.

Type every root, separated by commas. A double sign is fine: 2 ± 3i means 2 + 3i and 2 - 3i, and ±2i means 2i and -2i. Roots may also be left with the square root of a negative number, as (-5 ± sqrt(-3))/2.

Use of imaginaries

We have explained the laws of addition, subtraction, multiplication, and division for imaginary (and complex) numbers, and have made some use of them. It is largely because imaginaries obey these laws that we call them numbers. For it must be admitted that we cannot count objects with imaginary numbers. Nor can we state by means of them our age, our weight, or our balance at the bank. It should be remembered, however, that we can do none of these things with negative numbers. We may have a group of objects, books for example, whose number is 5; but no group of objects exists whose number is -5, or -3, or any negative number whatever. If it is asked, "How then can negative numbers and imaginary numbers have any practical use?" the answer is this: They have a practical use because the solution of equations forces them into our calculations, and when the necessary operations have been performed upon them, the result can frequently be interpreted.