This week, we will continue the topic of how to simplify rational expressions, started last week. We will also discuss solving quadratic equations and rational equations.
As we learned last week, the first step when performing operations on rational expressions is to factor polynomials. When rational expressions are added or subtracted, compute the LCD so that a single denominator is used. Finally, cancel any common factors to simplify the result of the operation. Radicals are similar to exponentials and follow the same mathematical principles.
As I noted last week in the list of definitions, an equation is a statement that two expressions are equal. To solve an equation, manipulate both expressions in the same way (addition, subtraction, multiplication, division, and exponentiation) in order to isolate the target variable on one side.
When you manipulate the expressions, beware of introducing false solutions by using zero in division or multiplication. We also must check the result to verify that it makes sense in the problem.
In a classic problem, there is an industrial party pack of 120 appetizers to be consumed at an office party. If two extra guests show up, everyone gets two fewer appetizers. How many guests were originally invited?
Formula 1.
Since the LCD of (x+2) and x is x(x+2) we can multiply both sides by (x)*(x+2). (the middle of Formula 1 shows how the right was derived).
Formula 2.
Canceling, we get
Formula 3. 120x = (120 – 2x)(x + 2) = 240 + 116x – 2x²
Subtracting (116x – 2x² + 240) from left and right sides, we get
Formula 4. 2x² + 4x - 240 = 0
Factoring gives us:
Formula 5. 2(x+12)(x-10) = 0
So the solutions are 10 and -12. Both of these solutions fit into formula 1; however, even the smallest party has a positive number of guests. Only the solution of ten invited guests makes sense.
As we saw in the transition from formula 2 to formula 3 above, an equation with a simple fraction on both sides can be simplified by multiplying both sides by the LCD. To make things even easier, we don't need the lowest common denominator, any common denominator will work. In fact, we can use the product of the two denominators in the equation.
If A = C
B D
We can multiply both sides by BD to get the equation AD = BC. This method, called the cross product, was once considered the final topic in learning arithmetic.
One of the basic principles of trigonometry is that if the angles of a triangle are the same, then the ratio of their sides is also identical. There is a great introduction to trigonometry at http://wright.nasa.gov/airplane/trig.html, but the principal of proportional triangles can be applied independently of trigonometry.
In a classic problem, a student wants to measure the height of a tree. The upper branches are too weak to support his weight, so he can't climb to the top and drop a line. Fortunately, the tree is in a clearing and the weather is sunny. He measures the length of the tree's shadow, 43 cubits (I said that this is a classic problem). At the same time, his friend measures the length of the shadow of a vertical one cubit standard; it is 48 finger-breadths (exactly 0.8 cubits) long. This is enough information to find the height of the tree. The triangle formed by the tree and its shadow has the same angle (the elevation of the sun) as the triangle formed by the one cubit standard and its shadow.
tree_height = 1 cubit
43 cubits 0.8 cubits
Dropping the cubit in the numerator and denominator and then cross multiplying, we get tree_height*0.8 = 43 cubits. Dividing both sides by 0.8, we get 53.75 cubits or 53 cubits and 45 finger-breadths.
Galileo discovered that a pendulum (a weight on a string) always takes the same amount of time to move back and forth even when the amount of swing changes. This discovery led to the invention of the grandfather clock. According to World Book, http://worldbook.bigchalk.com/421060.htm, the period of a string pendulum varies with the square root of the length of the string. A weight on a one meter string takes about one second to go from left to right and back to its original position.
In a sample problem, a chandelier hanging from the ceiling of the Paris opera house swung back and forth 25 times during a five minute aria. If the chandelier was 3 meters from the floor, how high was the ceiling? The period of the swing was 5 minutes (300 seconds) divided by 20, or 12 seconds. This period is 6 times that of a one meter pendulum, so the length of the cord of the chandelier is 36 meters. If we include the height of the chandelier, the ceiling is 39 meters high.
According to Kepler's third law the time that a satellite takes to complete an orbit is proportional to the height (from the center of the earth) to the 1.5th power. An old problem was to calculate the altitude of a geosynchronous orbit (one with a 24 hour period), based on the 240,000 mile altitude and 28 day period of the moon.
The solution is 240,000 / 282/3 = 26028 miles.
Under what situations would one or more solutions of a rational equation be unacceptable? Can you give examples?
(Read
Section 10.2 before answering.)
How would you solve the
equation (x² + 19x – 144)=0? Is it easier now than it was
last week?
How would you solve the equation (x² + 19x +
144)=0?
What is the altitude of a satellite which takes 90 minutes to complete its orbit? Is there a minimum time for a satellite to orbit the earth?