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# One-Variable Second-Order Equations for Physics Help (page 2)

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## Plugging In

Examine this equation once again:

−3 x 2 − 4 x − 2 = 0

Here, the coefficients are

a = −3

b = −4

c = −2

Plugging these numbers into the quadratic formula yields

x = {4 ± [(−4) 2 − (4 × −3 × −2)] 1/2 }/(2 × −3)

= 4 ± (16 − 24) 1/2 /−6

= 4 ± (−8) 1/2 /−6

We are confronted with the square root of −8 in the solution. This a “non-real” number.

## Those “nonreal” Numbers

Mathematicians symbolize the square root of −1, called the unit imaginary number , by using the lowercase italic letter i . Scientists and engineers more often symbolize it using the letter j , and henceforth, that is what we will do.

Any imaginary number can be obtained by multiplying j by some real number q . The real number q is customarily written after j if q is positive or zero. If q happens to be a negative real number, then the absolute value of q is written after − j . Examples of imaginary numbers are j 3, − j 5, j 2.787, and − j π.

The set of imaginary numbers can be depicted along a number line, just as can the real numbers. In a sense, the real-number line and the imaginary-number line are identical twins. As is the case with human twins, these two number lines, although they look similar, are independent. The sets of imaginary and real numbers have one value, zero, in common. Thus

j 0 = 0

A complex number consists of the sum of some real number and some imaginary number. The general form for a complex number k is

k = p + jq

where p and q are real numbers.

Mathematicians, scientists, and engineers all denote the set of complex numbers by placing the real-number and imaginary-number lines at right angles to each other, intersecting at zero. The result is a rectangular coordinate plane (Fig. 1-5). Every point on this plane corresponds to a unique complex number; every complex number corresponds to a unique point on the plane.

Fig. 1-5 . The complex numbers can be depicted graphically as points on a plane, defined by two number lines at right angles.

x = 4 ± j 8 1/2 /−6

Now that you know a little about complex numbers, you might want to examine the preceding solution and simplify it. Remember that it contains (−8) 1/2 . An engineer or physicist would write this as j 8 1/2 , so the solution to the quadratic is

## Back To “reality”

Now look again at this equation:

4 x 2 + 3 x − 5 = 0

Here, the coefficients are

a = 4

b = 3

c = −5

Plugging these numbers into the quadratic formula yields

x = {−3 ± [3 2 − (4 × 4 × −5)] 1/2 }/(2 × 4)

= −3 ± (9 + 80) 1/2 /8

= −3 ± (89) 1/2 /8

The square root of 89 is a real number but a messy one. When expressed in decimal form, it is nonrepeating and nonterminating. It can be approximated but never written out precisely. To four significant digits, its value is 9.434. Thus

x ≈ −6 ± 9.434/8

If you want to work this solution out to obtain two plain numbers without any addition, subtraction, or division operations in it, go ahead. However, it’s more important that you understand the process by which this solution is obtained. If you are confused on this issue, you’re better off reviewing the last several sections again and not bothering with arithmetic that any calculator can do for you mindlessly.

Practice problems for these concepts can be found at:  Equations, Formulas, And Vectors for Physics Practice Test

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