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Solve these cubic equations as a two-by-two system: y = 5x 3 + 3x 2 + 5x + 7 and y = 2x 3 + x 2 + 2x + 5

Clocks In electronics, the term clock refers to a circuit that generates pulses at high speed and at precise intervals It sets the tempo for the operation of digital devices In a computer, the clock acts like a metronome for the microprocessor Clock speeds are measured and expressed in hertz (Hz), kilohertz (kHz), megahertz (MHz), or gigahertz (GHz)

The whole range of electromagnetic frequencies or wavelengths is called the electromagnetic spectrum Theoretically there is no limit to how low or high the frequency can be, nor, correspondingly, to how long or short the wavelength can be The most common electromagnetic wavelengths range from about 106 m, or 1000 km, to around 10 12 m, or a trillionth of a meter Scientists use a logarithmic scale to depict the electromagnetic spectrum A simplified rendition is shown in Fig 26-15, labeled for wavelength in meters To find the frequencies in megahertz, divide 300 by the wavelength shown For frequencies in hertz, use 300,000,000 instead of 300 For kilohertz, use 300,000; for gigahertz, use 0300

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Solving this problem requires some keen intuition, a lot of trial and error, or both We are lucky in one respect, at least: These equations are already functions of x, so we have no morphing to do We can directly mix the right sides to get 5x 3 + 3x 2 + 5x + 7 = 2x 3 + x 2 + 2x + 5 Let s subtract the entire right side of this equation, as a single quantity, from both sides That changes each term on the left and sets the right side equal to 0, giving us 3x 3 + 2x 2 + 3x + 2 = 0

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Radio waves fall into a subset of this spectrum, at frequencies between approximately 9 kHz and 1000 GHz This corresponds to wavelengths of 33 km and 03 mm The radio spectrum, which includes television and microwaves, is blown up in Fig 26- 15, and is labeled for frequency To find wavelengths, use the conversion formulas Just be sure you have the decimal point in the right place with respect to the 3

Table 26-3

Now here s the trick! If we spend enough time playing around with this, we ll find that it can be factored into (3x + 2)(x 2 + 1) = 0 When we set the first binomial equal to 0, we get 3x + 2 = 0 which resolves to x = 2/3 When we set the second binomial equal to 0, we get x2 + 1 = 0 Subtracting 1 from both sides gives us x 2 = 1 This resolves to x = j or x = j We have found three roots to the cubic we got by mixing: x = 2/3 or x = j or x = j We can substitute these roots into either of the original cubics Let s use the second one For x = 2/3, we have y = 2x 3 + x 2 + 2x + 5 = 2 ( 2/3)3 + ( 2/3)2 + 2 ( 2/3) + 5 = 2 ( 8/27) + 4/9 + ( 4/3) + 5 = 16/27 + 12/27 + ( 36/27) + 135/27 = 95/27 Our first solution is (x , y) = ( 2/3, 95/27) For x = j, we have y = 2x 3 + x 2 + 2x + 5 = 2j 3 + j 2 + 2j + 5 = 2( j ) + ( 1) + j 2 + 5 = j 2 + ( 1) + j 2 + 5 = j 2 + j 2 + ( 1) + 5 =0+4 =4 Our second solution is (x, y) = (j, 4) For x = j, we have y = 2x 3 + x 2 + 2x + 5 = 2( j )3 + ( j )2 + 2( j ) + 5 = 2j + ( 1) + ( j 2) + 5.

Data can be transmitted over various different media The most common are cable, radio, satellite links, and fiberoptics Cable, radio/TV, and satellite communications use the radio-frequency spectrum Fiberoptics uses infrared or visible light energy

= j 2 + ( 1) + ( j 2) + 5 = j 2 + j 2 + ( 1) + 5 =0+4 =4 Our third solution is (x , y) = ( j, 4)

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